To the Student
As you begin, you may feel anxious about the number of theorems, definitions,
procedures, and equations. You may wonder if you can learn it all in time. Don’t
worry—your concerns are normal. This textbook was written with you in mind. If you
attend class, work hard, and read and study this text, you will build the knowledge
and skills you need to be successful. Here’s how you can use the text to your benefit.
Read Carefully
When you get busy, it’s easy to skip reading and go right to the problems. Don’t . . .
the text has a large number of examples and clear explanations to help you break
down the mathematics into easy-to-understand steps. Reading will provide you with
a clearer understanding, beyond simple memorization. Read before class (not after)
so you can ask questions about anything you didn’t understand. You’ll be amazed at
how much more you’ll get out of class if you do this.
Use the Features
I use many different methods in the classroom to communicate. Those methods,
when incorporated into the text, are called “features.” The features serve many
purposes, from providing timely review of material you learned before (just when
you need it) to providing organized review sessions to help you prepare for quizzes
and tests. Take advantage of the features and you will master the material.
To make this easier, we’ve provided a brief guide to getting the most from this
text. Refer to “Prepare for Class,” “Practice,” and “Review” on the following three
pages. Spend fifteen minutes reviewing the guide and familiarizing yourself with the
features by flipping to the page numbers provided. Then, as you read, use them. This
is the best way to make the most of your text.
Please do not hesitate to contact us, through Pearson Education, with any
questions, comments, or suggestions for improving this text. I look forward to
hearing from you, and good luck with all of your studies.
Best Wishes!
Feature
Description
Benefit
Page
Every Chapter Opener begins with . . .
Chapter- Opening Each chapter begins with a discussion of
a topic of current interest and ends with a
Topic & Project
The Project lets you apply what you
learned to solve a problem related to the
topic.
402
The projects allow for the integration of
spreadsheet technology that you will need
to be a productive member of the
workforce.
The projects give you an opportunity to
collaborate and use mathematics to deal
with issues of current interest.
503
Each section begins with a list of objectives.
Objectives also appear in the text where
the objective is covered.
These focus your studying by emphasizing
what’s most important and where to find it.
423
PREPARING FOR
THIS SECTION
Most sections begin with a list of key
concepts to review with page numbers.
Ever forget what you’ve learned? This
feature highlights previously learned material
to be used in this section. Review it, and
you’ll always be prepared to move forward.
423
Now Work the
‘Are You Prepared?’
Problems
Problems that assess whether you have the Not sure you need the Preparing for This 423, 434
prerequisite knowledge for the upcoming
Section review? Work the ‘Are You
section.
Prepared?’ problems. If you get one wrong,
you’ll know exactly what you need to
review and where to review it!
Now Work
These follow most examples and direct
you to a related exercise.
related project.
Internet-Based
Projects
Every Section begins with . . .
Learning Objectives
2
Sections contain . . .
PROBLEMS
WARNING
Warnings are provided in the text.
We learn best by doing. You’ll solidify
your understanding of examples if you try
a similar problem right away, to be sure
you understand what you’ve just read.
These point out common mistakes and
help you to avoid them.
430, 435
456
418, 443
These graphing utility activities foreshadow
a concept or solidify a concept just
presented.
You will obtain a deeper and more
intuitive understanding of theorems and
definitions.
In Words
These provide alternative descriptions of
select definitions and theorems.
Does math ever look foreign to you? This
feature translates math into plain English.
Calculus
These appear next to information essential
for the study of calculus.
Pay attention–if you spend extra time
now, you’ll do better later!
These examples provide “how-to”
instruction by offering a guided, step-by-step
approach to solving a problem.
With each step presented on the left and
the mathematics displayed on the right,
you can immediately see how each step
is employed.
334
These examples and problems require you
to build a mathematical model from either
a verbal description or data. The homework
Model It! problems are marked by purple
headings.
It is rare for a problem to come in the
form “Solve the following equation.”
Rather, the equation must be developed
based on an explanation of the problem.
These problems require you to develop
models that will allow you to describe the
problem mathematically and suggest a
solution to the problem.
447, 475
Exploration and
Seeing the Concept
SHOWCASE EXAMPLES
Model It! Examples
and Problems
440
205, 407,
431
Feature
Description
Benefit
Page
‘Are You Prepared?’
Problems
These assess your retention of the
prerequisite material you’ll need. Answers
are given at the end of the section exercises.
This feature is related to the Preparing for
This Section feature.
Do you always remember what you’ve
learned? Working these problems is
the best way to find out. If you get one
wrong, you’ll know exactly what you
need to review and where to review it!
434, 440
Concepts and
Vocabulary
These short-answer questions, mainly
Fill-in-the-Blank, Multiple-Choice and
True/False items, assess your understanding
of key definitions and concepts in the
current section.
It is difficult to learn math without
knowing the language of mathematics.
These problems test your understanding
of the formulas and vocabulary.
434
Skill Building
Correlated with section examples, these
problems provide straightforward practice.
It’s important to dig in and develop your
skills. These problems provide you with
ample opportunity to do so.
434–436
Mixed Practice
These problems offer comprehensive
assessment of the skills learned in the
section by asking problems that relate to
more than one concept or objective. These
problems may also require you to utilize
skills learned in previous sections.
Learning mathematics is a building
process. Many concepts are interrelated.
These problems help you see how
mathematics builds on itself and also see
how the concepts tie together.
436–437
Applications and
Extensions
These problems allow you to apply your
skills to real-world problems. They also
allow you to extend concepts learned in
the section.
You will see that the material learned
within the section has many uses in
everyday life.
437–439
Explaining Concepts: “Discussion and Writing” problems are
colored red. They support class
Discussion and
discussion, verbalization of mathematical
Writing
To verbalize an idea, or to describe it
clearly in writing, shows real understanding.
These problems nurture that understanding.
Many are challenging, but you’ll get out
what you put in.
439
NEW!
Retain Your
Knowledge
These problems allow you to practice
content learned earlier in the course.
Remembering how to solve all the
different kinds of problems that
you encounter throughout the course
is difficult. This practice helps you
remember.
439
Now Work
Many examples refer you to a related
homework problem. These related
problems are marked by a pencil and
orange numbers.
If you get stuck while working problems,
look for the closest Now Work problem,
and refer to the related example to
see if it helps.
Every chapter concludes with a
comprehensive list of exercises to pratice.
Use the list of objectives to determine the
objective and examples that correspond
to the problems.
Work these problems to ensure that you 499–501
understand all the skills and concepts of
the chapter. Think of it as a comprehensive
review of the chapter.
ideas, and writing and research projects.
PROBLEMS
Review Exercises
432, 435,
436
Feature
Description
Benefit
Page
The Chapter Review at the end of each chapter contains . . .
Things to Know
A detailed list of important theorems,
formulas, and definitions from the chapter.
Review these and you’ll know the most
important material in the chapter!
You Should Be
Able to . . .
Contains a complete list of objectives
by section, examples that illustrate the
objective, and practice exercises that test
your understanding of the objective.
Do the recommended exercises and you’ll 498–499
have mastered the key material. If you
get something wrong, go back and work
through the example listed and try again.
Review Exercises
These provide comprehensive review and Practice makes perfect. These problems 499–501
practice of key skills, matched to the Learning combine exercises from all sections,
giving you a comprehensive review in one
Objectives for each section.
place.
Chapter Test
About 15–20 problems that can be taken Be prepared. Take the sample practice
as a Chapter Test. Be sure to take the Chapter test under test conditions. This will get you
ready for your instructor’s test. If you get a
Test under test conditions—no notes!
problem wrong, you can watch the Chapter
Test Prep Video.
Cumulative Review
These problem sets appear at the end of
each chapter, beginning with Chapter 2.
They combine problems from previous
chapters, providing an ongoing cumulative
review. When you use them in conjunction
with the Retain Your Knowledge problems,
you will be ready for the final exam.
These problem sets are really important. 502–503
Completing them will ensure that you are
not forgetting anything as you go. This will
go a long way toward keeping you primed
for the final exam.
Chapter Projects
The Chapter Projects apply to what
you’ve learned in the chapter. Additional
projects are available on the Instructor’s
Resource Center (IRC).
The Chapter Projects give you an opportunity 503–504
to apply what you’ve learned in the chapter
to the opening topic. If your instructor
allows, these make excellent opportunities
to work in a group, which is often the best
way of learning math.
Internet-Based
Projects
In selected chapters, a Web-based project These projects give you an opportunity to
is given.
collaborate and use mathematics to deal
with issues of current interest by using the
Internet to research and collect data.
497–498
502
503
Achieve Your Potential
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in MyMathLab® to ensure you have many resources
to help you achieve success in mathematics - and beyond!
The MyMathLab features described here will help you:
‡ Review math skills and concepts you may have forgotten
‡ Retain new concepts as you move through your math course
‡ Develop skills that will help with your transition to college
Adaptive Study Plan
The Study Plan will help you study
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Your performance and activity are
assessed continually in real time,
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Skills for Success
The Skills for Success Modules
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college, online learning, time
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Additional content is provided to
help with the development of
professional skills such as resume
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Getting Ready
Are you frustrated when you
know you learned a math
concept in the past, but you
can’t quite remember the skill
when it’s time to use it?
Don’t worry!
The author has included
Getting Ready material so you
can brush up on forgotten
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Then, a personalized homework assignment provides additional
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Retain Your Knowledge
As you work through your math
course, these MyMathLab®
exercises support ongoing
review to help you maintain
HVVHQWLDOVNLOOV
The ability to recall important
math concepts as you continually
acquire new mathematical skills
will help you be successful in
this math course and in your
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Algebra & Trigonometry
Tenth Edition
Michael Sullivan
Chicago State University
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The student edition of this text has been cataloged as follows:
Library of Congress Cataloging-in-Publication Data
Sullivan, Michael, 1942Algebra & trigonometry / Michael Sullivan, Chicago State University. -- Tenth edition.
pages cm.
ISBN 978-0-321-99859-0
1. Algebra--Textbooks. 2. Algebra--Study and teaching (Higher) 3. Trigonometry--Textbooks.
4. Trigonometry--Study and teaching (Higher) I. Title. II. Title: Algebra and trigonometry.
QA154.3.S73 2016
512’ .13--dc23
2014021731
Copyright © 2016 by Pearson Education, Inc. or its affiliates. All Rights Reserved. Printed in the
United States of America. This publication is protected by copyright, and permission should be
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PEARSON, ALWAYS LEARNING, and MYMATHLAB are exclusive trademarks in the
U.S. and/or other countries owned by Pearson Education, Inc. or its affiliates.
1 2 3 4 5 6 7 8 9 10—CRK—17 16 15 14
www.pearsonhighered.com
ISBN-10: 0-321-99859-6
ISBN-13: 978-0-321-99859-0
Contents
Three Distinct Series
R
xviii
The Contemporary Series
xix
Preface to the Instructor
xx
Resources for Success
xxiv
Applications Index
xxvi
Review
1
R.1 Real Numbers
2
8PSLXJUI4FUTr$MBTTJGZ/VNCFSTr&WBMVBUF/VNFSJDBM&YQSFTTJPOT
r8PSLXJUI1SPQFSUJFTPG3FBM/VNCFST
R.2 Algebra Essentials
17
(SBQI*OFRVBMJUJFTr'JOE%JTUBODFPOUIF3FBM/VNCFS-JOFr&WBMVBUF
"MHFCSBJD&YQSFTTJPOTr%FUFSNJOFUIF%PNBJOPGB7BSJBCMFr6TFUIF
-BXTPG&YQPOFOUTr&WBMVBUF4RVBSF3PPUTr6TFB$BMDVMBUPSUP&WBMVBUF
&YQPOFOUTr6TF4DJFOUJGJD/PUBUJPO
R.3
Geometry Essentials
30
6TFUIF1ZUIBHPSFBO5IFPSFNBOE*UT$POWFSTFr,OPX(FPNFUSZ
'PSNVMBTr6OEFSTUBOE$POHSVFOU5SJBOHMFTBOE4JNJMBS5SJBOHMFT
R.4 Polynomials
39
3FDPHOJ[F.POPNJBMTr3FDPHOJ[F1PMZOPNJBMTr"EEBOE4VCUSBDU
1PMZOPNJBMTr.VMUJQMZ1PMZOPNJBMTr,OPX'PSNVMBTGPS4QFDJBM1SPEVDUT
r%JWJEF1PMZOPNJBMT6TJOH-POH%JWJTJPOr8PSLXJUI1PMZOPNJBMTJO5XP
Variables
R.5 Factoring Polynomials
49
Factor the Difference of Two Squares and the Sum and Difference of Two
$VCFTr'BDUPS1FSGFDU4RVBSFTr'BDUPSB4FDPOE%FHSFF
Polynomial: x2 + Bx + Cr'BDUPSCZ(SPVQJOHr'BDUPSB4FDPOE%FHSFF
Polynomial: Ax2 + Bx + C, A ≠ 1r$PNQMFUFUIF4RVBSF
R.6 Synthetic Division
58
Divide Polynomials Using Synthetic Division
R.7 Rational Expressions
62
3FEVDFB3BUJPOBM&YQSFTTJPOUP-PXFTU5FSNTr.VMUJQMZBOE%JWJEF
3BUJPOBM&YQSFTTJPOTr"EEBOE4VCUSBDU3BUJPOBM&YQSFTTJPOTr6TFUIF
-FBTU$PNNPO.VMUJQMF.FUIPEr4JNQMJGZ$PNQMFY3BUJPOBM&YQSFTTJPOT
R.8 nth Roots; Rational Exponents
73
Work with nUI3PPUTr4JNQMJGZ3BEJDBMTr3BUJPOBMJ[F%FOPNJOBUPST
r4JNQMJGZ&YQSFTTJPOTXJUI3BUJPOBM&YQPOFOUT
1
Equations and Inequalities
1.1 Linear Equations
81
82
4PMWFB-JOFBS&RVBUJPOr4PMWF&RVBUJPOT5IBU-FBEUP-JOFBS&RVBUJPOT
r4PMWF1SPCMFNT5IBU$BO#F.PEFMFECZ-JOFBS&RVBUJPOT
1.2 Quadratic Equations
92
4PMWFB2VBESBUJD&RVBUJPOCZ'BDUPSJOHr4PMWFB2VBESBUJD&RVBUJPOCZ
$PNQMFUJOHUIF4RVBSFr4PMWFB2VBESBUJD&RVBUJPO6TJOHUIF2VBESBUJD
'PSNVMBr4PMWF1SPCMFNT5IBU$BO#F.PEFMFECZ2VBESBUJD&RVBUJPOT
vii
viii
CONTENTS
1.3 Complex Numbers; Quadratic Equations in the Complex
Number System
104
Add, Subtract, Multiply, and Divide Complex Numbers
r4PMWF2VBESBUJD&RVBUJPOTJOUIF$PNQMFY/VNCFS4ZTUFN
1.4 Radical Equations; Equations Quadratic in Form;
Factorable Equations
113
4PMWF3BEJDBM&RVBUJPOTr4PMWF&RVBUJPOT2VBESBUJDJO'PSNr4PMWF
Equations by Factoring
1.5 Solving Inequalities
119
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*OFRVBMJUJFTr4PMWF$PNCJOFE*OFRVBMJUJFT
1.6 Equations and Inequalities Involving Absolute Value
130
4PMWF&RVBUJPOT*OWPMWJOH"CTPMVUF7BMVFr4PMWF*OFRVBMJUJFT*OWPMWJOH
Absolute Value
1.7 Problem Solving: Interest, Mixture, Uniform Motion,
Constant Rate Job Applications
134
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*OUFSFTU1SPCMFNTr4PMWF.JYUVSF1SPCMFNTr4PMWF6OJGPSN.PUJPO
1SPCMFNTr4PMWF$POTUBOU3BUF+PC1SPCMFNT
2
Chapter Review
143
Chapter Test
147
Chapter Projects
147
Graphs
2.1 The Distance and Midpoint Formulas
149
150
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2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
157
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*OUFSDFQUTGSPNBO&RVBUJPOr5FTUBO&RVBUJPOGPS4ZNNFUSZXJUI3FTQFDU
to the x-Axis, the y"YJT BOEUIF0SJHJOr,OPX)PXUP(SBQI,FZ
Equations
2.3 Lines
167
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BOEUIF4MPQFr'JOEUIF&RVBUJPOPGB7FSUJDBM-JOFr6TFUIF1PJOUm4MPQF
'PSNPGB-JOF*EFOUJGZ)PSJ[POUBM-JOFTr'JOEUIF&RVBUJPOPGB-JOF
(JWFO5XP1PJOUTr8SJUFUIF&RVBUJPOPGB-JOFJO4MPQFm*OUFSDFQU'PSN
r*EFOUJGZUIF4MPQFBOEy*OUFSDFQUPGB-JOFGSPN*UT&RVBUJPOr(SBQI
-JOFT8SJUUFOJO(FOFSBM'PSN6TJOH*OUFSDFQUTr'JOE&RVBUJPOTPG1BSBMMFM
-JOFTr'JOE&RVBUJPOTPG1FSQFOEJDVMBS-JOFT
2.4 Circles
182
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r8PSLXJUIUIF(FOFSBM'PSNPGUIF&RVBUJPOPGB$JSDMF
2.5 Variation
188
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*OWFSTF7BSJBUJPOr$POTUSVDUB.PEFM6TJOH+PJOU7BSJBUJPOPS$PNCJOFE
Variation
Chapter Review
194
Chapter Test
196
Cumulative Review
196
Chapter Project
197
CONTENTS
3
Functions and Their Graphs
198
3.1 Functions
199
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'VODUJPOr'JOEUIF%JGGFSFODF2VPUJFOUPGB'VODUJPOr'JOEUIF%PNBJO
PGB'VODUJPO%FGJOFECZBO&RVBUJPOr'PSNUIF4VN %JGGFSFODF 1SPEVDU and Quotient of Two Functions
3.2 The Graph of a Function
214
*EFOUJGZUIF(SBQIPGB'VODUJPOr0CUBJO*OGPSNBUJPOGSPNPSBCPVUUIF
Graph of a Function
3.3 Properties of Functions
223
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0EE'VODUJPOTGSPNBO&RVBUJPOr6TFB(SBQIUP%FUFSNJOF8IFSFB
'VODUJPO*T*ODSFBTJOH %FDSFBTJOH PS$POTUBOUr6TFB(SBQIUP-PDBUF
-PDBM.BYJNBBOE-PDBM.JOJNBr6TFB(SBQIUP-PDBUFUIF"CTPMVUF
.BYJNVNBOEUIF"CTPMVUF.JOJNVNr6TFB(SBQIJOH6UJMJUZUP
Approximate Local Maxima and Local Minima and to Determine Where a
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of a Function
3.4 Library of Functions; Piecewise-defined Functions
237
(SBQIUIF'VODUJPOT-JTUFEJOUIF-JCSBSZPG'VODUJPOTr(SBQI
Piecewise-defined Functions
3.5 Graphing Techniques: Transformations
247
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6TJOH$PNQSFTTJPOTBOE4USFUDIFTr(SBQI'VODUJPOT6TJOH3FGMFDUJPOT
about the x-Axis and the y-Axis
3.6 Mathematical Models: Building Functions
260
Build and Analyze Functions
4
Chapter Review
266
Chapter Test
270
Cumulative Review
271
Chapter Projects
271
Linear and Quadratic Functions
273
4.1 Properties of Linear Functions and Linear Models
274
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'VODUJPOTr%FUFSNJOF8IFUIFSB-JOFBS'VODUJPO*T*ODSFBTJOH %FDSFBTJOH PS$POTUBOUr#VJME-JOFBS.PEFMTGSPN7FSCBM%FTDSJQUJPOT
4.2 Building Linear Models from Data
284
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BOE/POMJOFBS3FMBUJPOTr6TFB(SBQIJOH6UJMJUZUP'JOEUIF-JOF
of Best Fit
4.3 Quadratic Functions and Their Properties
290
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BOE"YJTPG4ZNNFUSZPGB2VBESBUJD'VODUJPOr(SBQIB2VBESBUJD
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Value of a Quadratic Function
4.4 Build Quadratic Models from Verbal Descriptions and from Data 302
#VJME2VBESBUJD.PEFMTGSPN7FSCBM%FTDSJQUJPOTr#VJME2VBESBUJD.PEFMT
from Data
ix
x
CONTENTS
4.5 Inequalities Involving Quadratic Functions
312
Solve Inequalities Involving a Quadratic Function
5
Chapter Review
315
Chapter Test
318
Cumulative Review
319
Chapter Projects
320
Polynomial and Rational Functions
5.1 Polynomial Functions and Models
321
322
*EFOUJGZ1PMZOPNJBM'VODUJPOTBOE5IFJS%FHSFFr(SBQI1PMZOPNJBM
'VODUJPOT6TJOH5SBOTGPSNBUJPOTr*EFOUJGZUIF3FBM;FSPTPGB1PMZOPNJBM
'VODUJPOBOE5IFJS.VMUJQMJDJUZr"OBMZ[FUIF(SBQIPGB1PMZOPNJBM
'VODUJPOr#VJME$VCJD.PEFMTGSPN%BUB
5.2 Properties of Rational Functions
343
'JOEUIF%PNBJOPGB3BUJPOBM'VODUJPOr'JOEUIF7FSUJDBM"TZNQUPUFT
PGB3BUJPOBM'VODUJPOr'JOEUIF)PSJ[POUBMPS0CMJRVF"TZNQUPUFPGB
Rational Function
5.3 The Graph of a Rational Function
353
"OBMZ[FUIF(SBQIPGB3BUJPOBM'VODUJPOr4PMWF"QQMJFE1SPCMFNT
Involving Rational Functions
5.4 Polynomial and Rational Inequalities
368
4PMWF1PMZOPNJBM*OFRVBMJUJFTr4PMWF3BUJPOBM*OFRVBMJUJFT
5.5 The Real Zeros of a Polynomial Function
375
6TFUIF3FNBJOEFSBOE'BDUPS5IFPSFNTr6TF%FTDBSUFT3VMFPG4JHOTUP
%FUFSNJOFUIF/VNCFSPG1PTJUJWFBOEUIF/VNCFSPG/FHBUJWF3FBM;FSPT
PGB1PMZOPNJBM'VODUJPOr6TFUIF3BUJPOBM;FSPT5IFPSFNUP-JTUUIF
1PUFOUJBM3BUJPOBM;FSPTPGB1PMZOPNJBM'VODUJPOr'JOEUIF3FBM;FSPTPG
B1PMZOPNJBM'VODUJPOr4PMWF1PMZOPNJBM&RVBUJPOTr6TFUIF5IFPSFNGPS
#PVOETPO;FSPTr6TFUIF*OUFSNFEJBUF7BMVF5IFPSFN
5.6 Complex Zeros; Fundamental Theorem of Algebra
390
6TFUIF$POKVHBUF1BJST5IFPSFNr'JOEB1PMZOPNJBM'VODUJPOXJUI
4QFDJGJFE;FSPTr'JOEUIF$PNQMFY;FSPTPGB1PMZOPNJBM'VODUJPO
6
Chapter Review
396
Chapter Test
399
Cumulative Review
399
Chapter Projects
400
Exponential and Logarithmic Functions
6.1 Composite Functions
402
403
'PSNB$PNQPTJUF'VODUJPOr'JOEUIF%PNBJOPGB$PNQPTJUF'VODUJPO
6.2 One-to-One Functions; Inverse Functions
411
%FUFSNJOF8IFUIFSB'VODUJPO*T0OFUP0OFr%FUFSNJOFUIF*OWFSTFPGB
'VODUJPO%FGJOFECZB.BQPSB4FUPG0SEFSFE1BJSTr0CUBJOUIF(SBQIPG
UIF*OWFSTF'VODUJPOGSPNUIF(SBQIPGUIF'VODUJPOr'JOEUIF*OWFSTFPGB
Function Defined by an Equation
6.3 Exponential Functions
&WBMVBUF&YQPOFOUJBM'VODUJPOTr(SBQI&YQPOFOUJBM'VODUJPOTr%FGJOF
the Number er4PMWF&YQPOFOUJBM&RVBUJPOT
423
CONTENTS
6.4 Logarithmic Functions
440
Change Exponential Statements to Logarithmic Statements and
-PHBSJUINJD4UBUFNFOUTUP&YQPOFOUJBM4UBUFNFOUTr&WBMVBUF-PHBSJUINJD
&YQSFTTJPOTr%FUFSNJOFUIF%PNBJOPGB-PHBSJUINJD'VODUJPOr(SBQI
-PHBSJUINJD'VODUJPOTr4PMWF-PHBSJUINJD&RVBUJPOT
6.5 Properties of Logarithms
452
8PSLXJUIUIF1SPQFSUJFTPG-PHBSJUINTr8SJUFB-PHBSJUINJD
&YQSFTTJPOBTB4VNPS%JGGFSFODFPG-PHBSJUINTr8SJUFB-PHBSJUINJD
&YQSFTTJPOBTB4JOHMF-PHBSJUINr&WBMVBUF-PHBSJUINT8IPTF#BTF*T
Neither 10 Nor e
6.6 Logarithmic and Exponential Equations
461
4PMWF-PHBSJUINJD&RVBUJPOTr4PMWF&YQPOFOUJBM&RVBUJPOTr4PMWF
Logarithmic and Exponential Equations Using a Graphing Utility
6.7 Financial Models
468
%FUFSNJOFUIF'VUVSF7BMVFPGB-VNQ4VNPG.POFZr$BMDVMBUF
&GGFDUJWF3BUFTPG3FUVSOr%FUFSNJOFUIF1SFTFOU7BMVFPGB-VNQ
4VNPG.POFZr%FUFSNJOFUIF3BUFPG*OUFSFTUPSUIF5JNF3FRVJSFEUP
Double a Lump Sum of Money
6.8 Exponential Growth and Decay Models; Newton’s Law;
Logistic Growth and Decay Models
478
Find Equations of Populations That Obey the Law of Uninhibited Growth
r'JOE&RVBUJPOTPG1PQVMBUJPOT5IBU0CFZUIF-BXPG%FDBZr6TF/FXUPOT
-BXPG$PPMJOHr6TF-PHJTUJD.PEFMT
6.9 Building Exponential, Logarithmic, and Logistic
Models from Data
489
#VJMEBO&YQPOFOUJBM.PEFMGSPN%BUBr#VJMEB-PHBSJUINJD.PEFMGSPN
%BUBr#VJMEB-PHJTUJD.PEFMGSPN%BUB
7
Chapter Review
497
Chapter Test
502
Cumulative Review
502
Chapter Projects
503
Trigonometric Functions
505
7.1 Angles and Their Measure
506
Convert between Decimal and Degree, Minute, Second Measures for
"OHMFTr'JOEUIF-FOHUIPGBO"SDPGB$JSDMFr$POWFSUGSPN
%FHSFFTUP3BEJBOTBOEGSPN3BEJBOTUP%FHSFFTr'JOEUIF"SFBPGB
4FDUPSPGB$JSDMFr'JOEUIF-JOFBS4QFFEPGBO0CKFDU5SBWFMJOHJO
Circular Motion
7.2 Right Triangle Trigonometry
519
'JOEUIF7BMVFTPG5SJHPOPNFUSJD'VODUJPOTPG"DVUF"OHMFTr6TF
'VOEBNFOUBM*EFOUJUJFTr'JOEUIF7BMVFTPGUIF3FNBJOJOH5SJHPOPNFUSJD
'VODUJPOT (JWFOUIF7BMVFPG0OFPG5IFNr6TFUIF$PNQMFNFOUBSZ"OHMF
Theorem
7.3 Computing the Values of Trigonometric Functions
of Acute Angles
p
Find the Exact Values of the Trigonometric Functions of
= 45°r'JOE
4
p
p
= 30° and
= 60°
the Exact Values of the Trigonometric Functions of
6
3
r6TFB$BMDVMBUPSUP"QQSPYJNBUFUIF7BMVFTPGUIF5SJHPOPNFUSJD
'VODUJPOTPG"DVUF"OHMFTr.PEFMBOE4PMWF"QQMJFE1SPCMFNT*OWPMWJOH
Right Triangles
531
xi
xii
CONTENTS
7.4 Trigonometric Functions of Any Angle
543
'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD'VODUJPOTGPS"OZ"OHMFr6TF
Coterminal Angles to Find the Exact Value of a Trigonometric Function
r%FUFSNJOFUIF4JHOTPGUIF5SJHPOPNFUSJD'VODUJPOTPGBO"OHMFJOB(JWFO
2VBESBOUr'JOEUIF3FGFSFODF"OHMFPGBO"OHMFr6TFB3FGFSFODF"OHMF
UP'JOEUIF&YBDU7BMVFPGB5SJHPOPNFUSJD'VODUJPOr'JOEUIF&YBDU7BMVFT
of the Trigonometric Functions of an Angle, Given Information about the
Functions
7.5 Unit Circle Approach; Properties of the Trigonometric
Functions
553
Find the Exact Values of the Trigonometric Functions Using the Unit
$JSDMFr,OPXUIF%PNBJOBOE3BOHFPGUIF5SJHPOPNFUSJD'VODUJPOTr6TF
Periodic Properties to Find the Exact Values of the Trigonometric Functions
r6TF&WFOm0EE1SPQFSUJFTUP'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD
Functions
7.6 Graphs of the Sine and Cosine Functions
564
7.7 Graphs of the Tangent, Cotangent, Cosecant, and
Secant Functions
579
7.8 Phase Shift; Sinusoidal Curve Fitting
587
Graph Functions of the Form y = A sin (vx) 6TJOH5SBOTGPSNBUJPOTr(SBQI
Functions of the Form y = A cos (vx)6TJOH5SBOTGPSNBUJPOTr%FUFSNJOF
UIF"NQMJUVEFBOE1FSJPEPG4JOVTPJEBM'VODUJPOTr(SBQI4JOVTPJEBM
'VODUJPOT6TJOH,FZ1PJOUTr'JOEBO&RVBUJPOGPSB4JOVTPJEBM(SBQI
Graph Functions of the Form y = A tan (vx) + B and y = A cot (vx) + B
r(SBQI'VODUJPOTPGUIF'PSNy = A csc (vx) + B and y = A sec (vx) + B
Graph Sinusoidal Functions of the Form y = A sin (vx - f) + B
r#VJME4JOVTPJEBM.PEFMTGSPN%BUB
8
Chapter Review
597
Chapter Test
603
Cumulative Review
603
Chapter Projects
604
Analytic Trigonometry
8.1 The Inverse Sine, Cosine, and Tangent Functions
606
607
'JOEUIF&YBDU7BMVFPGBO*OWFSTF4JOF'VODUJPOr'JOEBO"QQSPYJNBUF
7BMVFPGBO*OWFSTF4JOF'VODUJPOr6TF1SPQFSUJFTPG*OWFSTF'VODUJPOT
UP'JOE&YBDU7BMVFTPG$FSUBJO$PNQPTJUF'VODUJPOTr'JOEUIF*OWFSTF
'VODUJPOPGB5SJHPOPNFUSJD'VODUJPOr4PMWF&RVBUJPOT*OWPMWJOH*OWFSTF
Trigonometric Functions
8.2 The Inverse Trigonometric Functions (Continued)
620
Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, and
5BOHFOU'VODUJPOTr%FGJOFUIF*OWFSTF4FDBOU $PTFDBOU BOE$PUBOHFOU
'VODUJPOTr6TFB$BMDVMBUPSUP&WBMVBUFsec -1x, csc -1x, and cot -1xr8SJUF
a Trigonometric Expression as an Algebraic Expression
8.3 Trigonometric Equations
625
4PMWF&RVBUJPOT*OWPMWJOHB4JOHMF5SJHPOPNFUSJD'VODUJPOr4PMWF
5SJHPOPNFUSJD&RVBUJPOT6TJOHB$BMDVMBUPSr4PMWF5SJHPOPNFUSJD&RVBUJPOT
2VBESBUJDJO'PSNr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOH'VOEBNFOUBM
*EFOUJUJFTr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOHB(SBQIJOH6UJMJUZ
8.4 Trigonometric Identities
6TF"MHFCSBUP4JNQMJGZ5SJHPOPNFUSJD&YQSFTTJPOTr&TUBCMJTI*EFOUJUJFT
635
CONTENTS
8.5 Sum and Difference Formulas
xiii
643
6TF4VNBOE%JGGFSFODF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF4VNBOE
%JGGFSFODF'PSNVMBTUP&TUBCMJTI*EFOUJUJFTr6TF4VNBOE%JGGFSFODF
'PSNVMBT*OWPMWJOH*OWFSTF5SJHPOPNFUSJD'VODUJPOTr4PMWF5SJHPOPNFUSJD
Equations Linear in Sine and Cosine
8.6 Double-angle and Half-angle Formulas
655
6TF%PVCMFBOHMF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF%PVCMFBOHMF'PSNVMBT
UP&TUBCMJTI*EFOUJUJFTr6TF)BMGBOHMF'PSNVMBTUP'JOE&YBDU7BMVFT
8.7 Product-to-Sum and Sum-to-Product Formulas
665
&YQSFTT1SPEVDUTBT4VNTr&YQSFTT4VNTBT1SPEVDUT
9
Chapter Review
669
Chapter Test
672
Cumulative Review
673
Chapter Projects
674
Applications of Trigonometric Functions
9.1 Applications Involving Right Triangles
675
676
4PMWF3JHIU5SJBOHMFTr4PMWF"QQMJFE1SPCMFNT
9.2 The Law of Sines
682
4PMWF4""PS"4"5SJBOHMFTr4PMWF44"5SJBOHMFTr4PMWF"QQMJFE1SPCMFNT
9.3 The Law of Cosines
692
4PMWF4"45SJBOHMFTr4PMWF4445SJBOHMFTr4PMWF"QQMJFE1SPCMFNT
9.4 Area of a Triangle
699
'JOEUIF"SFBPG4"45SJBOHMFTr'JOEUIF"SFBPG4445SJBOHMFT
9.5 Simple Harmonic Motion; Damped Motion; Combining Waves
705
#VJMEB.PEFMGPSBO0CKFDUJO4JNQMF)BSNPOJD.PUJPOr"OBMZ[F
4JNQMF)BSNPOJD.PUJPOr"OBMZ[FBO0CKFDUJO%BNQFE.PUJPO
r(SBQIUIF4VNPG5XP'VODUJPOT
10
Chapter Review
714
Chapter Test
716
Cumulative Review
717
Chapter Projects
718
Polar Coordinates; Vectors
10.1 Polar Coordinates
719
720
1MPU1PJOUT6TJOH1PMBS$PPSEJOBUFTr$POWFSUGSPN1PMBS$PPSEJOBUFTUP
3FDUBOHVMBS$PPSEJOBUFTr$POWFSUGSPN3FDUBOHVMBS$PPSEJOBUFTUP1PMBS
$PPSEJOBUFTr5SBOTGPSN&RVBUJPOTCFUXFFO1PMBSBOE3FDUBOHVMBS'PSNT
10.2 Polar Equations and Graphs
729
Identify and Graph Polar Equations by Converting to Rectangular
&RVBUJPOTr5FTU1PMBS&RVBUJPOTGPS4ZNNFUSZr(SBQI1PMBS&RVBUJPOT
by Plotting Points
10.3 The Complex Plane; De Moivre’s Theorem
1MPU1PJOUTJOUIF$PNQMFY1MBOFr$POWFSUB$PNQMFY/VNCFSCFUXFFO
3FDUBOHVMBS'PSNBOE1PMBS'PSNr'JOE1SPEVDUTBOE2VPUJFOUTPG
$PNQMFY/VNCFSTJO1PMBS'PSNr6TF%F.PJWSFT5IFPSFNr'JOE
Complex Roots
744
xiv
CONTENTS
10.4 Vectors
752
(SBQI7FDUPSTr'JOEB1PTJUJPO7FDUPSr"EEBOE4VCUSBDU7FDUPST
"MHFCSBJDBMMZr'JOEB4DBMBS.VMUJQMFBOEUIF.BHOJUVEFPGB7FDUPS
r'JOEB6OJU7FDUPSr'JOEB7FDUPSGSPN*UT%JSFDUJPOBOE.BHOJUVEF
r.PEFMXJUI7FDUPST
10.5 The Dot Product
766
'JOEUIF%PU1SPEVDUPG5XP7FDUPSTr'JOEUIF"OHMFCFUXFFO5XP7FDUPST
r%FUFSNJOF8IFUIFS5XP7FDUPST"SF1BSBMMFMr%FUFSNJOF8IFUIFS5XP
7FDUPST"SF0SUIPHPOBMr%FDPNQPTFB7FDUPSJOUP5XP0SUIPHPOBM7FDUPST
r$PNQVUF8PSL
11
Chapter Review
773
Chapter Test
776
Cumulative Review
776
Chapter Projects
777
Analytic Geometry
11.1 Conics
778
779
Know the Names of the Conics
11.2 The Parabola
780
"OBMZ[F1BSBCPMBTXJUI7FSUFYBUUIF0SJHJOr"OBMZ[F1BSBCPMBTXJUI
Vertex at (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH1BSBCPMBT
11.3 The Ellipse
789
"OBMZ[F&MMJQTFTXJUI$FOUFSBUUIF0SJHJOr"OBMZ[F&MMJQTFTXJUI
Center at (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH&MMJQTFT
11.4 The Hyperbola
799
"OBMZ[F)ZQFSCPMBTXJUI$FOUFSBUUIF0SJHJOr'JOEUIF"TZNQUPUFTPG
B)ZQFSCPMBr"OBMZ[F)ZQFSCPMBTXJUI$FOUFSBU(h, k)r4PMWF"QQMJFE
Problems Involving Hyperbolas
11.5 Rotation of Axes; General Form of a Conic
812
*EFOUJGZB$POJDr6TFB3PUBUJPOPG"YFTUP5SBOTGPSN&RVBUJPOT
r"OBMZ[FBO&RVBUJPO6TJOHB3PUBUJPOPG"YFTr*EFOUJGZ$POJDTXJUIPVU
a Rotation of Axes
11.6 Polar Equations of Conics
820
"OBMZ[FBOE(SBQI1PMBS&RVBUJPOTPG$POJDTr$POWFSUUIF1PMBS
Equation of a Conic to a Rectangular Equation
11.7 Plane Curves and Parametric Equations
826
(SBQI1BSBNFUSJD&RVBUJPOTr'JOEB3FDUBOHVMBS&RVBUJPOGPSB
$VSWF%FGJOFE1BSBNFUSJDBMMZr6TF5JNFBTB1BSBNFUFSJO1BSBNFUSJD
&RVBUJPOTr'JOE1BSBNFUSJD&RVBUJPOTGPS$VSWFT%FGJOFECZ
Rectangular Equations
12
Chapter Review
838
Chapter Test
841
Cumulative Review
841
Chapter Projects
842
Systems of Equations and Inequalities
12.1 Systems of Linear Equations: Substitution and Elimination
4PMWF4ZTUFNTPG&RVBUJPOTCZ4VCTUJUVUJPOr4PMWF4ZTUFNTPG&RVBUJPOT
CZ&MJNJOBUJPOr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH
843
844
CONTENTS
xv
5XP7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT
$POUBJOJOH5XP7BSJBCMFTr4PMWF4ZTUFNTPG5ISFF&RVBUJPOT$POUBJOJOH
5ISFF7BSJBCMFTr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH
5ISFF7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT
Containing Three Variables
12.2 Systems of Linear Equations: Matrices
859
8SJUFUIF"VHNFOUFE.BUSJYPGB4ZTUFNPG-JOFBS&RVBUJPOTr8SJUFUIF
4ZTUFNPG&RVBUJPOTGSPNUIF"VHNFOUFE.BUSJYr1FSGPSN3PX0QFSBUJPOT
POB.BUSJYr4PMWFB4ZTUFNPG-JOFBS&RVBUJPOT6TJOH.BUSJDFT
12.3 Systems of Linear Equations: Determinants
874
&WBMVBUFCZ%FUFSNJOBOUTr6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG
5XP&RVBUJPOT$POUBJOJOH5XP7BSJBCMFTr&WBMVBUFCZ%FUFSNJOBOUT
r6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG5ISFF&RVBUJPOT$POUBJOJOH5ISFF
7BSJBCMFTr,OPX1SPQFSUJFTPG%FUFSNJOBOUT
12.4 Matrix Algebra
884
'JOEUIF4VNBOE%JGGFSFODFPG5XP.BUSJDFTr'JOE4DBMBS.VMUJQMFTPGB
.BUSJYr'JOEUIF1SPEVDUPG5XP.BUSJDFTr'JOEUIF*OWFSTFPGB.BUSJY
Solve a System of Linear Equations Using an Inverse Matrix
12.5 Partial Fraction Decomposition
901
P
Decompose , Where Q Has Only Nonrepeated Linear Factors
Q
P
P
r%FDPNQPTF , Where Q)BT3FQFBUFE-JOFBS'BDUPSTr%FDPNQPTF ,
Q
Q
P
Where Q)BTB/POSFQFBUFE*SSFEVDJCMF2VBESBUJD'BDUPSr%FDPNQPTF ,
Q
Where Q Has a Repeated Irreducible Quadratic Factor
12.6 Systems of Nonlinear Equations
909
4PMWFB4ZTUFNPG/POMJOFBS&RVBUJPOT6TJOH4VCTUJUVUJPOr4PMWFB
System of Nonlinear Equations Using Elimination
12.7 Systems of Inequalities
918
(SBQIBO*OFRVBMJUZr(SBQIB4ZTUFNPG*OFRVBMJUJFT
12.8 Linear Programming
925
4FU6QB-JOFBS1SPHSBNNJOH1SPCMFNr4PMWFB-JOFBS1SPHSBNNJOH
Problem
13
Chapter Review
932
Chapter Test
936
Cumulative Review
937
Chapter Projects
937
Sequences; Induction; the Binomial Theorem
13.1 Sequences
939
940
8SJUFUIF'JSTU4FWFSBM5FSNTPGB4FRVFODFr8SJUFUIF5FSNTPGB
4FRVFODF%FGJOFECZB3FDVSTJWF'PSNVMBr6TF4VNNBUJPO/PUBUJPO
r'JOEUIF4VNPGB4FRVFODF
13.2 Arithmetic Sequences
950
%FUFSNJOF8IFUIFSB4FRVFODF*T"SJUINFUJDr'JOEB'PSNVMBGPSBO
"SJUINFUJD4FRVFODFr'JOEUIF4VNPGBO"SJUINFUJD4FRVFODF
13.3 Geometric Sequences; Geometric Series
%FUFSNJOF8IFUIFSB4FRVFODF*T(FPNFUSJDr'JOEB'PSNVMBGPSB
(FPNFUSJD4FRVFODFr'JOEUIF4VNPGB(FPNFUSJD4FRVFODF
r%FUFSNJOF8IFUIFSB(FPNFUSJD4FSJFT$POWFSHFTPS%JWFSHFTr4PMWF
Annuity Problems
956
xvi
CONTENTS
13.4 Mathematical Induction
967
Prove Statements Using Mathematical Induction
13.5 The Binomial Theorem
971
n
Evaluate a b r6TFUIF#JOPNJBM5IFPSFN
j
14
Chapter Review
977
Chapter Test
979
Cumulative Review
980
Chapter Projects
981
Counting and Probability
14.1 Counting
982
983
'JOE"MMUIF4VCTFUTPGB4FUr$PVOUUIF/VNCFSPG&MFNFOUTJOB4FU
r4PMWF$PVOUJOH1SPCMFNT6TJOHUIF.VMUJQMJDBUJPO1SJODJQMF
14.2 Permutations and Combinations
988
Solve Counting Problems Using Permutations Involving n Distinct
0CKFDUTr4PMWF$PVOUJOH1SPCMFNT6TJOH$PNCJOBUJPOTr4PMWF$PVOUJOH
Problems Using Permutations Involving n Nondistinct Objects
14.3 Probability
997
$POTUSVDU1SPCBCJMJUZ.PEFMTr$PNQVUF1SPCBCJMJUJFTPG&RVBMMZ-JLFMZ
0VUDPNFTr'JOE1SPCBCJMJUJFTPGUIF6OJPOPG5XP&WFOUTr6TFUIF
Complement Rule to Find Probabilities
Appendix
Chapter Review
1007
Chapter Test
1009
Cumulative Review
1010
Chapter Projects
1010
Graphing Utilities
A1
A.1 The Viewing Rectangle
A1
A.2 Using a Graphing Utility to Graph Equations
A3
A.3 Using a Graphing Utility to Locate Intercepts and Check for
Symmetry
A5
A.4 Using a Graphing Utility to Solve Equations
A6
A.5 Square Screens
A8
A.6 Using a Graphing Utility to Graph Inequalities
A9
A.7 Using a Graphing Utility to Solve Systems of Linear Equations
A9
A.8 Using a Graphing Utility to Graph a Polar Equation
A11
A.9 Using a Graphing Utility to Graph Parametric Equations
A11
Answers
AN1
Credits
Index
C1
I1
For the family
Katy (Murphy) and Pat
Shannon, Patrick, Ryan
Mike and Yola
Michael, Kevin, Marissa
Dan and Sheila
Maeve, Sean, Nolan
Colleen (O’Hara) and Bill
Kaleigh, Billy, Timmy
Three Distinct Series
Students have different goals, learning styles, and levels of preparation. Instructors
have different teaching philosophies, styles, and techniques. Rather than write one
series to fit all, the Sullivans have written three distinct series. All share the same
goal—to develop a high level of mathematical understanding and an appreciation
for the way mathematics can describe the world around us. The manner of reaching
that goal, however, differs from series to series.
Contemporary Series, Tenth Edition
The Contemporary Series is the most traditional in approach yet modern in its
treatment of precalculus mathematics. Graphing utility coverage is optional and can
be included or excluded at the discretion of the instructor: College Algebra, Algebra
& Trigonometry, Trigonometry: A Unit Circle Approach, Precalculus.
Enhanced with Graphing Utilities Series,
Sixth Edition
This series provides a thorough integration of graphing utilities into topics, allowing
students to explore mathematical concepts and encounter ideas usually studied in
later courses. Using technology, the approach to solving certain problems differs
from the Contemporary Series, while the emphasis on understanding concepts and
building strong skills does not: College Algebra, Algebra & Trigonometry, Precalculus.
Concepts through Functions Series,
Third Edition
This series differs from the others, utilizing a functions approach that serves as the
organizing principle tying concepts together. Functions are introduced early in
various formats. This approach supports the Rule of Four, which states that functions
are represented symbolically, numerically, graphically, and verbally. Each chapter
introduces a new type of function and then develops all concepts pertaining to that
particular function. The solutions of equations and inequalities, instead of being
developed as stand-alone topics, are developed in the context of the underlying
functions. Graphing utility coverage is optional and can be included or excluded
at the discretion of the instructor: College Algebra; Precalculus, with a Unit Circle
Approach to Trigonometry; Precalculus, with a Right Triangle Approach to
Trigonometry.
xviii
The Contemporary Series
College Algebra, Tenth Edition
This text provides a contemporary approach to college algebra, with three chapters
of review material preceding the chapters on functions. Graphing calculator usage
is provided, but is optional. After completing this book, a student will be adequately
prepared for trigonometry, finite mathematics, and business calculus.
Algebra & Trigonometry, Tenth Edition
This text contains all the material in College Algebra, but also develops the
trigonometric functions using a right triangle approach and showing how it
relates to the unit circle approach. Graphing techniques are emphasized, including a
thorough discussion of polar coordinates, parametric equations, and conics using
polar coordinates. Graphing calculator usage is provided, but is optional. After
completing this book, a student will be adequately prepared for finite mathematics,
business calculus, and engineering calculus.
Precalculus, Tenth Edition
This text contains one review chapter before covering the traditional precalculus
topic of functions and their graphs, polynomial and rational functions, and
exponential and logarithmic functions. The trigonometric functions are introduced
using a unit circle approach and showing how it relates to the right triangle
approach. Graphing techniques are emphasized, including a thorough discussion of
polar coordinates, parametric equations, and conics using polar coordinates. Graphing
calculator usage is provided, but is optional. The final chapter provides an
introduction to calculus, with a discussion of the limit, the derivative, and the
integral of a function. After completing this book, a student will be adequately
prepared for finite mathematics, business calculus, and engineering calculus.
Trigonometry: a Unit Circle Approach, Tenth Edition
This text, designed for stand-alone courses in trigonometry, develops the trigonometric
functions using a unit circle approach and showing how it relates to the right
triangle approach. Graphing techniques are emphasized, including a thorough
discussion of polar coordinates, parametric equations, and conics using polar
coordinates. Graphing calculator usage is provided, but is optional. After completing
this book, a student will be adequately prepared for finite mathematics, business
calculus, and engineering calculus.
xix
Preface to the Instructor
A
s a professor of mathematics at an urban public
university for 35 years, I understand the varied
needs of algebra and trigonometry students. Students
range from being underprepared, with little mathematical
background and a fear of mathematics, to being highly
prepared and motivated. For some, this is their final course
in mathematics. For others, it is preparation for future
mathematics courses. I have written this text with both
groups in mind.
A tremendous benefit of authoring a successful series is
the broad-based feedback I receive from teachers and students
who have used previous editions. I am sincerely grateful for
their support. Virtually every change to this edition is the
result of their thoughtful comments and suggestions. I hope
that I have been able to take their ideas and, building upon a
successful foundation of the ninth edition, make this series an
even better learning and teaching tool for students and teachers.
Features in the Tenth Edition
A descriptive list of the many special features of
Algebra & Trigonometry can be found on the endpapers in
the front of this text.
This list places the features in their proper context, as
building blocks of an overall learning system that has been
carefully crafted over the years to help students get the most
out of the time they put into studying. Please take the time to
review this and to discuss it with your students at the beginning
of your course. My experience has been that when students
utilize these features, they are more successful in the course.
New to the Tenth Edition
r Retain Your Knowledge This new category of problems
in the exercise set are based on the article “To Retain
New Learning, Do the Math” published in the Edurati
Review. In this article, Kevin Washburn suggests that “the
more students are required to recall new content or skills,
the better their memory will be.” It is frustrating when
students cannot recall skills learned earlier in the course.
To alleviate this recall problem, we have created “Retain
Your Knowledge” problems. These are problems considered
to be “final exam material” that students can use to maintain
their skills. All the answers to these problems appear in the
back of the text, and all are programmed in MyMathLab.
r Guided Lecture Notes Ideal for online, emporium/
redesign courses, inverted classrooms, or traditional
lecture classrooms. These lecture notes help students
take thorough, organized, and understandable notes
as they watch the Author in Action videos. They ask
students to complete definitions, procedures, and
examples based on the content of the videos and text.
In addition, experience suggests that students learn by
doing and understanding the why/how of the concept or
xx
r
r
r
r
property.Therefore, many sections will have an exploration
activity to motivate student learning. These explorations
introduce the topic and/or connect it to either a real-world
application or a previous section. For example, when the
vertical-line test is discussed in Section 3.2, after the
theorem statement, the notes ask the students to explain
why the vertical-line test works by using the definition
of a function. This challenge helps students process the
information at a higher level of understanding.
Illustrations Many of the figures now have captions to
help connect the illustrations to the explanations in the
body of the text.
TI Screen Shots In this edition we have replaced all
the screen shots from the ninth edition with screen shots
using TI-84Plus C. These updated screen shots help
students visualize concepts clearly and help make stronger
connections between equations, data, and graphs in full
color.
Chapter Projects, which apply the concepts of each
chapter to a real-world situation, have been enhanced
to give students an up-to-the-minute experience. Many
projects are new and Internet-based, requiring the student
to research information online in order to solve problems.
Exercise Sets All the exercises in the text have been
reviewed and analyzed for this edition, some have been
removed, and new ones have been added. All time-sensitive
problems have been updated to the most recent information
available. The problem sets remain classified according
to purpose.
The ‘Are You Prepared?’ problems have been
improved to better serve their purpose as a just-in-time
review of concepts that the student will need to apply in
the upcoming section.
The Concepts and Vocabulary problems have been
expanded and now include multiple-choice exercises.
Together with the fill-in-the-blank and True/False
problems, these exercises have been written to serve as
reading quizzes.
Skill Building problems develop the student’s
computational skills with a large selection of exercises
that are directly related to the objectives of the section.
Mixed Practice problems offer a comprehensive assessment
of skills that relate to more than one objective. Often
these require skills learned earlier in the course.
Applications and Extensions problems have been
updated. Further, many new application-type exercises
have been added, especially ones involving information
and data drawn from sources the student will recognize,
to improve relevance and timeliness.
The Explaining Concepts: Discussion and Writing
exercises have been improved and expanded to provide
more opportunity for classroom discussion and group
projects.
PREFACE
xxi
New to this edition, Retain Your Knowledge exercises
consist of a collection of four problems in each exercise set
that are based on material learned earlier in the course.
They serve to keep information that has already been
learned “fresh” in the mind of the student. Answers to all
these problems appear in the Student Edition.
The Review Exercises in the Chapter Review have
been streamlined, but they remain tied to the clearly
expressed objectives of the chapter. Answers to all these
problems appear in the Student Edition.
r Annotated Instructor’s Edition As a guide, the author’s
suggestions for homework assignments are indicated by
a blue underscore below the problem number. These
problems are assignable in the MyMathLab as part of a
“Ready-to-Go” course.
Chapter R Review
This chapter consists of review material. It may be used as the
first part of the course or later as a just-in-time review when
the content is required. Specific references to this chapter
occur throughout the text to assist in the review process.
Content Changes in the Tenth Edition
Chapter 3 Functions and Their Graphs
Perhaps the most important chapter. Section 3.6 is optional.
r Section 3.1 The objective Find the Difference Quotient
of a Function has been added.
r Section 5.1 The subsection Behavior of the Graph of a
1PMZOPNJBM'VODUJPO/FBSB;FSPIBTCFFOSFNPWFE
r Section 5.3 A subsection has been added that discusses
the role of multiplicity of the zeros of the denominator
of a rational function as it relates to the graph near a
vertical asymptote.
r Section 5.5 The objective Use Descartes’ Rule of Signs
has been included.
r Section 5.5 5IF UIFPSFN #PVOET PO UIF ;FSPT PG B
Polynomial Function is now based on the traditional
method of using synthetic division.
Chapter 4 Linear and Quadratic Functions
Topic selection depends on your syllabus. Sections 4.2
and 4.4 may be omitted without loss of continuity.
Using the Tenth Edition Effectively
with Your Syllabus
To meet the varied needs of diverse syllabi, this text
contains more content than is likely to be covered in an
Algebra & Trigonometry course.As the chart illustrates, this text
has been organized with flexibility of use in mind.Within a given
chapter, certain sections are optional (see the details that follow
the figure below) and can be omitted without loss of continuity.
1
R
4
5
6
11.111.4 12 14
7
13
8
9
10.110.3
10.410.5
Chapter 2 Graphs
This chapter lays the foundation for functions. Section 2.5
is optional.
Chapter 5 Polynomial and Rational Functions
Topic selection depends on your syllabus.
Chapter 6 Exponential and Logarithmic Functions
4FDUJPOT m GPMMPX JO TFRVFODF 4FDUJPOT and 6.9 are optional.
Chapter 7 Trigonometric Functions
Section 7.8 may be omitted in a brief course.
Chapter 8 Analytic Trigonometry
Sections 8.2, 8.6, and 8.8 may be omitted in a brief course.
Chapter 9 Applications of Trigonometric Functions
Sections 9.4 and 9.5 may be omitted in a brief course.
Chapter 10 Polar Coordinates; Vectors
4FDUJPOTmBOE4FDUJPOTmBSFJOEFQFOEFOU
and may be covered separately.
Chapter 11 Analytic Geometry
4FDUJPOT m GPMMPX JO TFRVFODF 4FDUJPOT and 11.7 are independent of each other, but each requires
4FDUJPOTm
Chapter 12 Systems of Equations and Inequalities
4FDUJPOTmNBZCFDPWFSFEJOBOZPSEFS CVUFBDI
requires Section 12.1. Section 12.8 requires Section 12.7.
2
3
Chapter 1 Equations and Inequalities
Primarily a review of Intermediate Algebra topics, this
material is a prerequisite for later topics. The coverage of
complex numbers and quadratic equations with a negative
discriminant is optional and may be postponed or skipped
entirely without loss of continuity.
11.511.7
Chapter 13 Sequences; Induction; The Binomial
Theorem
5IFSF BSF UISFF JOEFQFOEFOU QBSUT 4FDUJPOT m
Section 13.4; and Section 13.5.
Chapter 14 Counting and Probability
The sections follow in sequence.
xxii
PREFACE
Acknowledgments
Textbooks are written by authors, but evolve from an idea
to final form through the efforts of many people. It was
Don Dellen who first suggested this text and series to me.
Don is remembered for his extensive contributions to
publishing and mathematics.
Thanks are due to the following people for their
assistance and encouragement to the preparation of this
edition:
r From Pearson Education: Anne Kelly for her substantial
contributions, ideas, and enthusiasm; Dawn Murrin,
for her unmatched talent at getting the details right;
Joseph Colella for always getting the reviews and pages
to me on time; Peggy McMahon for directing the always
difficult production process; Rose Kernan for handling
James Africh, College of DuPage
Steve Agronsky, Cal Poly State University
Gererdo Aladro, Florida International
University
Grant Alexander, Joliet Junior College
Dave Anderson, South Suburban College
Richard Andrews, Florida A&M University
Joby Milo Anthony, University of Central
Florida
James E. Arnold, University of
Wisconsin-Milwaukee
Adel Arshaghi, Center for Educational Merit
Carolyn Autray, University of West Georgia
Agnes Azzolino, Middlesex County College
Wilson P. Banks, Illinois State University
Sudeshna Basu, Howard University
Dale R. Bedgood, East Texas State University
Beth Beno, South Suburban College
Carolyn Bernath, Tallahassee Community
College
Rebecca Berthiaume, Edison State College
William H. Beyer, University of Akron
Annette Blackwelder, Florida State University
Richelle Blair, Lakeland Community College
Kevin Bodden, Lewis and Clark College
Jeffrey Boerner, University of Wisconsin-Stout
Barry Booten, Florida Atlantic University
Larry Bouldin, Roane State Community
College
Bob Bradshaw, Ohlone College
Trudy Bratten, Grossmont College
Tim Bremer, Broome Community College
Tim Britt, Jackson State Community College
Michael Brook, University of Delaware
Joanne Brunner, Joliet Junior College
Warren Burch, Brevard Community College
Mary Butler, Lincoln Public Schools
Melanie Butler, West Virginia University
Jim Butterbach, Joliet Junior College
William J. Cable, University of
Wisconsin-Stevens Point
Lois Calamia, Brookdale Community College
Jim Campbell, Lincoln Public Schools
Roger Carlsen, Moraine Valley Community
College
Elena Catoiu, Joliet Junior College
Mathews Chakkanakuzhi, Palomar College
Tim Chappell, Penn Valley Community College
John Collado, South Suburban College
Alicia Collins, Mesa Community College
Nelson Collins, Joliet Junior College
Rebecca Connell, Troy University
Jim Cooper, Joliet Junior College
Denise Corbett, East Carolina University
liaison between the compositor and author; Peggy
Lucas for her genuine interest in marketing this text;
Chris Hoag for her continued support and genuine
interest; Greg Tobin for his leadership and commitment
to excellence; and the Pearson Math and Science Sales
team, for their continued confidence and personal
support of our texts.
r Accuracy checkers: C. Brad Davis, who read the entire
manuscript and accuracy checked answers. His
attention to detail is amazing; Timothy Britt, for creating
the Solutions Manuals and accuracy checking
answers.
Finally, I offer my grateful thanks to the dedicated
users and reviewers of my texts, whose collective
insights form the backbone of each textbook revision.
Carlos C. Corona, San Antonio College
Theodore C. Coskey, South Seattle
Community College
Rebecca Connell, Troy University
Donna Costello, Plano Senior High School
Paul Crittenden, University of Nebraska at
Lincoln
John Davenport, East Texas State University
Faye Dang, Joliet Junior College
Antonio David, Del Mar College
Stephanie Deacon, Liberty University
Duane E. Deal, Ball State University
Jerry DeGroot, Purdue North Central
Timothy Deis, University of WisconsinPlatteville
Joanna DelMonaco, Middlesex Community
College
Vivian Dennis, Eastfield College
Deborah Dillon, R. L. Turner High School
Guesna Dohrman, Tallahassee Community
College
Cheryl Doolittle, Iowa State University
Karen R. Dougan, University of Florida
Jerrett Dumouchel, Florida Community
College at Jacksonville
Louise Dyson, Clark College
Paul D. East, Lexington Community College
Don Edmondson, University of Texas-Austin
Erica Egizio, Joliet Junior College
Jason Eltrevoog, Joliet Junior College
Christopher Ennis, University of Minnesota
Kathy Eppler, Salt Lake Community College
Ralph Esparza, Jr., Richland College
Garret J. Etgen, University of Houston
Scott Fallstrom, Shoreline Community College
Pete Falzone, Pensacola Junior College
Arash Farahmand, Skyline College
W.A. Ferguson, University of Illinois-Urbana/
Champaign
Iris B. Fetta, Clemson University
Mason Flake, student at Edison Community
College
Timothy W. Flood, Pittsburg State University
Robert Frank, Westmoreland County
Community College
Merle Friel, Humboldt State University
Richard A. Fritz, Moraine Valley
Community College
Dewey Furness, Ricks College
Mary Jule Gabiou, North Idaho College
Randy Gallaher, Lewis and Clark College
Tina Garn, University of Arizona
Dawit Getachew, Chicago State University
Wayne Gibson, Rancho Santiago College
Loran W. Gierhart, University of Texas at San
Antonio and Palo Alto College
Robert Gill, University of Minnesota Duluth
Nina Girard, University of Pittsburgh at
Johnstown
Sudhir Kumar Goel, Valdosta State University
Adrienne Goldstein, Miami Dade College,
Kendall Campus
Joan Goliday, Sante Fe Community College
Lourdes Gonzalez, Miami Dade College,
Kendall Campus
Frederic Gooding, Goucher College
Donald Goral, Northern Virginia Community
College
Sue Graupner, Lincoln Public Schools
Mary Beth Grayson, Liberty University
Jennifer L. Grimsley, University of Charleston
Ken Gurganus, University of North Carolina
James E. Hall, University of Wisconsin-Madison
Judy Hall, West Virginia University
Edward R. Hancock, DeVry Institute of
Technology
Julia Hassett, DeVry Institute, Dupage
Christopher Hay-Jahans, University of South
Dakota
Michah Heibel, Lincoln Public Schools
LaRae Helliwell, San Jose City College
Celeste Hernandez, Richland College
Gloria P. Hernandez, Louisiana State
University at Eunice
Brother Herron, Brother Rice High School
Robert Hoburg, Western Connecticut State
University
Lynda Hollingsworth, Northwest Missouri
State University
Deltrye Holt, Augusta State University
Charla Holzbog, Denison High School
Lee Hruby, Naperville North High School
Miles Hubbard, St. Cloud State University
Kim Hughes, California State College-San
Bernardino
Stanislav, Jabuka, University of Nevada, Reno
Ron Jamison, Brigham Young University
Richard A. Jensen, Manatee Community
College
Glenn Johnson, Middlesex Community College
Sandra G. Johnson, St. Cloud State University
Tuesday Johnson, New Mexico State
University
Susitha Karunaratne, Purdue University North
Central
Moana H. Karsteter, Tallahassee Community
College
Donna Katula, Joliet Junior College
PREFACE
Arthur Kaufman, College of Staten Island
Thomas Kearns, North Kentucky University
+BDL,FBUJOH .BTTBTPJU$PNNVOJUZ$PMMFHF
Shelia Kellenbarger, Lincoln Public Schools
Rachael Kenney, North Carolina State
University
+PIO#,MBTTFO /PSUI*EBIP$PMMFHF
Debra Kopcso, Louisiana State University
Lynne Kowski, Raritan Valley Community
College
Yelena Kravchuk, University of Alabama at
Birmingham
Ray S. Kuan, Skyline College
Keith Kuchar, Manatee Community College
Tor Kwembe, Chicago State University
-JOEB+,ZMF 5BSSBOU$PVOUSZ+S$PMMFHF
H.E. Lacey, Texas A & M University
Harriet Lamm, Coastal Bend College
+BNFT-BQQ 'PSU-FXJT$PMMFHF
Matt Larson, Lincoln Public Schools
Christopher Lattin, Oakton Community College
+VMJB-FEFU -PVTJBOB4UBUF6OJWFSTJUZ
Adele LeGere, Oakton Community College
Kevin Leith, University of Houston
+P"OO-FXJO &EJTPO$PMMFHF
+FGG-FXJT +PIOTPO$PVOUZ$PNNVOJUZ$PMMFHF
+BOJDF$-ZPO 5BMMBIBTTFF$PNNVOJUZ$PMMFHF
+FBO.D"SUIVS +PMJFU+VOJPS$PMMFHF
Virginia McCarthy, Iowa State University
Karla McCavit, Albion College
Michael McClendon, University of Central
Oklahoma
Tom McCollow, DeVry Institute of Technology
Marilyn McCollum, North Carolina State
University
+JMM.D(PXBO )PXBSE6OJWFSTJUZ
Will McGowant, Howard University
"OHFMB.D/VMUZ +PMJFU+VOJPS$PMMFHF
Laurence Maher, North Texas State University
+BZ".BMNTUSPN 0LMBIPNB$JUZ$PNNVOJUZ
College
Rebecca Mann, Apollo High School
Lynn Marecek, Santa Ana College
Sherry Martina, Naperville North High School
Alec Matheson, Lamar University
Nancy Matthews, University of Oklahoma
+BNFT.BYXFMM 0LMBIPNB4UBUF
University-Stillwater
Marsha May, Midwestern State University
+BNFT.D-BVHIMJO 8FTU$IFTUFS6OJWFSTJUZ
+VEZ.FDLMFZ +PMJFU+VOJPS$PMMFHF
David Meel, Bowling Green State University
Carolyn Meitler, Concordia University
Samia Metwali, Erie Community College
3JDI.FZFST +PMJFU+VOJPS$PMMFHF
Eldon Miller, University of Mississippi
+BNFT.JMMFS 8FTU7JSHJOJB6OJWFSTJUZ
Michael Miller, Iowa State University
Kathleen Miranda, SUNY at Old Westbury
Chris Mirbaha, The Community College of
Baltimore County
Val Mohanakumar, Hillsborough Community
College
Thomas Monaghan, Naperville North High
School
Miguel Montanez, Miami Dade College,
Wolfson Campus
Maria Montoya, Our Lady of the Lake
University
Susan Moosai, Florida Atlantic University
Craig Morse, Naperville North High School
Samad Mortabit, Metropolitan State University
Pat Mower, Washburn University
Tammy Muhs, University of Central Florida
A. Muhundan, Manatee Community College
+BOF.VSQIZ .JEEMFTFY$PNNVOJUZ$PMMFHF
Richard Nadel, Florida International University
Gabriel Nagy, Kansas State University
Bill Naegele, South Suburban College
Karla Neal, Lousiana State University
Lawrence E. Newman, Holyoke Community
College
Dwight Newsome, Pasco-Hernando
Community College
Denise Nunley, Maricopa Community Colleges
+BNFT/ZNBOO 6OJWFSTJUZPG5FYBT&M1BTP
Mark Omodt, Anoka-Ramsey Community
College
Seth F. Oppenheimer, Mississippi State
University
Leticia Oropesa, University of Miami
-JOEB1BEJMMB +PMJFU+VOJPS$PMMFHF
Sanja Pantic, University of Illinois at Chicago
&+BNFT1FBLF *PXB4UBUF6OJWFSTJUZ
Kelly Pearson, Murray State University
Dashamir Petrela, Florida Atlantic University
Philip Pina, Florida Atlantic University
Charlotte Pisors, Baylor University
Michael Prophet, University of Northern Iowa
Laura Pyzdrowski, West Virginia University
Carrie Quesnell, Weber State University
Neal C. Raber, University of Akron
5IPNBT3BEJO 4BO+PBRVJO%FMUB$PMMFHF
Aibeng Serene Radulovic, Florida Atlantic
University
Ken A. Rager, Metropolitan State College
Kenneth D. Reeves, San Antonio College
Elsi Reinhardt, Truckee Meadows Community
College
+PTF3FNFTBS .JBNJ%BEF$PMMFHF 8PMGTPO
Campus
+BOF3JOHXBME *PXB4UBUF6OJWFSTJUZ
Douglas F. Robertson, University of
Minnesota, MPLS
Stephen Rodi, Austin Community College
William Rogge, Lincoln Northeast High
School
Howard L. Rolf, Baylor University
Mike Rosenthal, Florida International
University
Phoebe Rouse, Lousiana State University
Edward Rozema, University of Tennessee at
Chattanooga
Dennis C. Runde, Manatee Community College
Alan Saleski, Loyola University of Chicago
4VTBO4BOENFZFS +BNFTUPXO$PNNVOJUZ
College
Brenda Santistevan, Salt Lake Community
College
Linda Schmidt, Greenville Technical College
Ingrid Scott, Montgomery College
A.K. Shamma, University of West Florida
;BDIFSZ4IBSPO 6OJWFSTJUZPG5FYBTBU4BO
Antonio
Martin Sherry, Lower Columbia College
Carmen Shershin, Florida International
University
5BUSBOB4IVCJO 4BO+PTF4UBUF6OJWFSTJUZ
Anita Sikes, Delgado Community College
Timothy Sipka, Alma College
xxiii
Charlotte Smedberg, University of Tampa
Lori Smellegar, Manatee Community College
Gayle Smith, Loyola Blakefield
Cindy Soderstrom, Salt Lake Community
College
Leslie Soltis, Mercyhurst College
+PIO4QFMMNBO 4PVUIXFTU5FYBT4UBUF
University
Karen Spike, University of North Carolina
Rajalakshmi Sriram, Okaloosa-Walton
Community College
Katrina Staley, North Carolina Agricultural
and Technical State University
Becky Stamper, Western Kentucky
University
+VEZ4UBWFS 'MPSJEB$PNNVOJUZ
College-South
Robin Steinberg, Pima Community College
Neil Stephens, Hinsdale South High School
Sonya Stephens, Florida A&M Univeristy
1BUSJDL4UFWFOT +PMJFU+VOJPS$PMMFHF
+PIO4VNOFS 6OJWFSTJUZPG5BNQB
Matthew TenHuisen, University of North
Carolina, Wilmington
Christopher Terry, Augusta State University
Diane Tesar, South Suburban College
Tommy Thompson, Brookhaven College
Martha K. Tietze, Shawnee Mission Northwest
High School
3JDIBSE+5POESB *PXB4UBUF6OJWFSTJUZ
Florentina Tone, University of West Florida
Suzanne Topp, Salt Lake Community College
Marilyn Toscano, University of Wisconsin,
Superior
Marvel Townsend, University of Florida
+JN5SVEOPXTLJ $BSSPMM$PMMFHF
3PCFSU5VTLFZ +PMJFU+VOJPS$PMMFHF
Mihaela Vajiac, Chapman University-Orange
+VMJB7BSCBMPX 5IPNBT/FMTPO$PNNVOJUZ
College-Leesville
Richard G. Vinson, University of South
Alabama
+PSHF7JPMB1SJPMJ 'MPSJEB"UMBOUJD6OJWFSTJUZ
Mary Voxman, University of Idaho
+FOOJGFS8BMTI %BZUPOB#FBDI$PNNVOJUZ
College
Donna Wandke, Naperville North High School
Timothy L.Warkentin, Cloud County
Community College
.FMJTTB+8BUUT 7JSHJOJB4UBUF6OJWFSTJUZ
Hayat Weiss, Middlesex Community College
Kathryn Wetzel, Amarillo College
Darlene Whitkenack, Northern Illinois
University
Suzanne Williams, Central Piedmont
Community College
Larissa Williamson, University of Florida
Christine Wilson, West Virginia University
Brad Wind, Florida International University
Anna Wiodarczyk, Florida International
University
Mary Wolyniak, Broome Community
College
Canton Woods, Auburn University
Tamara S. Worner, Wayne State College
Terri Wright, New Hampshire Community
Technical College, Manchester
"MFUIFJB;BNCFTJ 6OJWFSTJUZPG8FTU'MPSJEB
(FPSHF;B[J $IJDBHP4UBUF6OJWFSTJUZ
4UFWF;VSP +PMJFU+VOJPS$PMMFHF
Chicago State University
Resources for Success
Online Course (access code required)
MyMathLab delivers proven results in helping individual students succeed. It provides engaging
experiences that personalize, stimulate, and measure learning for each student. And it comes from an
experienced partner with educational expertise and an eye on the future. MyMathLab helps prepare
students and gets them thinking more conceptually and visually through the following features:
Adaptive Study Plan
The Study Plan makes studying more efficient and
effective for every student. Performance and activity
are assessed continually in real time. The data and
analytics are used to provide personalized content–
reinforcing concepts that target each student’s
strengths and weaknesses.
Getting Ready
Students refresh prerequisite topics through
assignable skill review quizzes and personalized
homework integrated in MyMathLab.
Video Assessment
Video assessment is tied to key Author in Action videos to check
students’ conceptual understanding of important math concepts.
Enhanced Graphing Functionality
New functionality within the graphing
utility allows graphing of 3-point quadratic
functions, 4-point cubic graphs, and
transformations in exercises.
Skills for Success Modules are integrated within the MyMathLab course to help
students succeed in collegiate courses and prepare for future professions.
Retain Your Knowledge These new exercises support ongoing review at the course
level and help students maintain essential skills.
xxiv
Instructor Resources
Student Resources
Additional resources can be downloaded from
www.pearsonhighered.com or hardcopy resources
can be ordered from your sales representative.
Additional resources to enhance student success:
Ready to Go MyMathLab® Course
Now it is even easier to get started with
MyMathLab. The Ready to Go MyMathLab course
option includes author-chosen preassigned
homework, integrated review, and more.
TestGen®
TestGen® (www.pearsoned.com/testgen) enables
instructors to build, edit, print, and administer tests
using a computerized bank of questions developed
to cover all the objectives of the text.
PowerPoint® Lecture Slides
Lecture Video
Author in Action videos are actual classroom
lectures with fully worked out examples presented
by Michael Sullivan, III. All video is assignable
within MyMathlab.
Chapter Test Prep Videos
Students can watch instructors work through
step-by-step solutions to all chapter test exercises
from the text. These are available in MyMathlab and
on YouTube.
Fully editable slides correlated with the text.
Student Solutions Manual
Annotated Instructor’s Edition
Provides detailed worked-out solutions to oddnumbered exercises.
Shorter answers are on the page beside the exercises.
Longer answers are in the back of the text.
Guided Lecture Notes
Includes additional examples and helpful teaching
tips, by section.
These lecture notes assist students in taking thorough,
organized, and understandable notes while watching
Author in Action videos. Students actively participate in
learning the how/why of important concepts through
explorations and activities. The Guided Lecture Notes
are available as PDF’s and customizable Word files in
MyMathLab. They can also be packaged with the text
and the MyMathLab access code.
Online Chapter Projects
Algebra Review
Additional projects that give students an opportunity
to apply what they learned in the chapter.
Four chapters of Intermediate Algebra review. Perfect
for a slower-paced course or for individual review.
Instructor Solutions Manual
Includes fully worked solutions to all exercises
in the text.
Mini Lecture Notes
xxv
Applications Index
Acoustics
Area. See also Geometry
amplifying sound, 500
loudness of sound, 451
loudspeaker, 712
tuning fork, 712, 713
whispering galleries, 795–796
of Bermuda Triangle, 703
under a curve, 619
of isosceles triangle, 664
of portion of rectangle outside of circle,
518
of sector of circle, 513, 516
of segment of circle, 715
of sidewalk, 531
for tethered dog to roam, 518
of windshield wiper sweep, 516
Aerodynamics
modeling aircraft motion, 777
Aeronautics
Challenger disaster, 488
Agriculture
farm management, 931
farm workers in U.S., 487
field enclosure, 916
grazing area for cow, 704
milk production, 494
minimizing cost, 931
removing stump, 765
watering a field, 102
Air travel
bearing of aircraft, 680
distance between two planes,
262
flight time and ticket price, 289
frequent flyer miles, 689
holding pattern, 633
parking at O’Hare International Airport,
245
revising a flight plan, 697
speed and direction of aircraft,
759, 763
Archaeology
age of ancient tools, 480–481
age of fossil, 486
age of tree, 486
date of prehistoric man’s death,
500
Architecture
brick staircase, 955, 979
Burj Khalifa building, 31
Flatiron Building, 703
floor design, 953–954, 979
football stadium seating, 955
mosaic design, 955, 979
Norman window, 37, 309
One World Trade Center, 541
parabolic arch, 309
racetrack design, 798
special window, 309, 317
stadium construction, 955
window design, 309
window dimensions, 102
xxvi
Art
fine decorative pieces, 541
framing a painting, 146
Astronomy
angle of elevation of Sun, 679
distance from Earth to its moon, 29
distances of planets from Sun, 949
International Space Station (ISS),
838
light-year, 29
planetary orbits
Earth, 798
elliptical, 798
Jupiter, 798
Mars, 798
Mercury, 825
Pluto, 798
Aviation
modeling aircraft motion, 777
orbital launches, 856
Biology
alcohol and driving, 447, 452
bacterial growth, 479–480, 493
E-coli, 235, 275
blood types, 987–988
bone length, 317–318
cricket chirp rate and temperature, 311
healing of wounds, 437, 451
maternal age versus Down syndrome,
290
muscle force, 764
yeast biomass as function of time, 492
Business
advertising, 318
automobile production, 409, 872
blending coffee, 141
candy bar size, 103
checkout lines, 1006
clothing store, 1008
commissions, 317
cookie orders, 935
copying machines, 146
cost
of can, 364, 367
of commodity, 410
of manufacturing, 29, 141, 374, 924–925
marginal, 301, 317
minimizing, 317, 931, 936
of printing, 337
of production, 234, 409, 899, 936
of transporting goods, 246
cost equation, 180, 192
cost function, 282
average, 217
demand
for candy, 192
demand equation, 317, 400
depreciation, 402
discount pricing, 91, 92, 410
drive-thru rate
at Burger King, 433
at Citibank, 437, 451
at McDonald’s, 438
equipment depreciation, 965
ethanol production, 493
expense computation, 142
farm workers in U.S., 487
Jiffy Lube’s car arrival rate, 437,
451
managing a meat market, 931
milk production, 494
mixing candy, 141
mixing nuts, 141
orange juice production, 872
precision ball bearings, 29
presale orders, 856
price markup, 91
of new car, 129
product design, 932
production scheduling, 931
product promotion, 181
profit, 899–900
maximizing, 929–930, 931–932
profit function, 213
rate of return on, 475
restaurant management, 856
revenue, 141, 301, 314–315
airline, 932
of clothing store, 889
daily, 301
from digital music, 259
from football seating, 966
maximizing, 301, 308
monthly, 301
theater, 857
revenue equation, 192
RV rental, 318–319
salary, 410, 955
gross, 212
increases in, 965, 979
Applications Index
sales
commission on, 128–129
of movie theater ticket, 844, 848–849,
856
net, 156
salvage value, 500
straight-line depreciation, 278–279, 282
supply and demand, 279–280, 282
tax, 374
theater attendance, 92
toy truck manufacturing, 924–925
transporting goods, 925
truck rentals, 180, 282–283
unemployment, 1009
wages
of car salesperson, 180
hourly, 89, 91
Calculus
absolute maximum/minimum, 228
area under a curve, 234, 260, 619
asymptotes, 346–347
average rate of change in, 230
bending wire into geometric shapes, 264
carrying a ladder around a corner,
633–634
cylinder inscribed in a cone, 264
cylinder inscribed in a sphere, 264
difference quotient, 205–206, 212, 439,
460, 654
express as single quotient in, 77
expressions with rational exponents in, 77
factoring in, 54, 58, 80
filling conical tank, 265
functions approximated by polynomial
functions, 342
increasing/decreasing functions,
225–226
inequalities involving absolute value,
134
infinite geometric series, 960
infinite limits, 332
Intermediate Value Theorem, 384, 385
limit notation, 580, 582, 960
limits at infinity, 332
local maxima/minima, 227
longest ladder carried around corner, 586,
633
maximizing projectile range, 633, 659, 664
maximizing rain gutter construction,
663–664
open box construction, 265
partial fraction decomposition, 902
quadratic equations, 99–100, 102
radians, 509
reducing expression to lowest
terms, 72
secant line, 231
Simpson’s rule, 309
Snell’s Law of Refraction, 634–635
tangent line, 703
the number e, 431–432
trigonometric functions, 657–658, 665
Carpentry. See also Construction
pitch, 182
Chemistry, 91
alpha particles, 811
decomposition reactions, 487
drug concentration, 366
gas laws, 193
mixing acids, 146
pH, 450
purity of gold, 142
radioactive decay, 486, 493–494,
500, 932
radioactivity from Chernobyl, 487
reactions, 309
salt solutions, 142, 146
sugar molecules, 142
volume of gas, 128
Combinatorics
airport codes, 989
binary codes, 1008
birthday permutations, 991, 995, 996,
1002–1003, 1007, 1008–1009
blouses and skirts combinations, 987
book arrangements, 995
box stacking, 995
code formation, 995
combination locks, 996
committee formation, 993, 995–996, 1008
Senate committees, 996
flag arrangement, 994, 1008
gender composition of children in family,
1000
letter codes, 989
license plate possibilities, 995, 1008, 1009
lining up people, 990, 995
number formation, 987, 995, 996, 1009
objects selection, 996
seating arrangements, 1008
shirts and ties combinations, 987
telephone numbers, 1008
two-symbol codewords, 986
word formation, 993–994, 996, 1009
Communications
cell phone towers, 495
installing cable TV, 265
international call plan, 283
phone charges, 282
satellite dish, 785–786, 787
spreading of rumors, 437, 451
surveillance satellites, 681
tablet service, 245
Touch-Tone phones, 668, 713
wireless data plan, 198, 234–235, 271–272
Computers and computing
comparing tablets, 103
graphics, 765, 900–901
households owning computers, 487
Internet searches, 112
iPod storage capacity, 283
xxvii
laser printers, 142
three-click rule, 900
website design, 900
website map, 900
Word users, 487
Construction
of border around a garden, 103
of border around a pool, 103
of box, 99–100, 102, 916
closed, 269
open, 265
of brick staircase, 979
of can, 398
of coffee can, 143
of cylindrical tube, 916
of enclosures
around garden, 142
around pond, 142
maximizing area of, 304, 308, 317
of fencing, 304, 308, 317, 916
minimum cost for, 366
of flashlight, 787
of headlight, 787
of highway, 540, 690, 715
installing cable TV, 265
painting a room, 586
patio dimensions, 103
pitch of roof, 680
of rain gutter, 309, 534–535, 663–664
of ramp, 689
access ramp, 181
of rectangular field enclosure, 308
of stadium, 309, 955
of steel drum, 367
of swimming pool, 37, 38
of swing set, 698
of tent, 702
TV dish, 787
vent pipe installation, 798
Cryptography
matrices in, 900
Decorating
Christmas tree, 32
Demographics
birth rate
age of mother and, 311
of unmarried women, 301
diversity index, 450
divorced population, 306–307
marital status, 988
mosquito colony growth, 486
population. See Population
rabbit colony growth, 948
Design
of awning, 691
of box with minimum surface
area, 367
of fine decorative pieces, 541
xxviii
Applications Index
of Little League Field, 518–519
of water sprinkler, 516
Direction
of aircraft, 759, 763
compass heading, 764
for crossing a river, 763
of fireworks display, 810
of lightning strikes, 810
of motorboat, 763
of swimmer, 775
Distance
Bermuda Triangle, 38
bicycle riding, 222
from Chicago to Honolulu, 619
circumference of Earth, 518
between cities, 511–512, 517
between Earth and Mercury, 691
between Earth and Venus, 691
from Earth to a star, 680
of explosion, 811
height
of aircraft, 689, 691
of bouncing ball, 965, 979
of bridge, 689
of building, 680
of cloud, 536–537
of CN Tower, 540
of Eiffel Tower, 540
of embankment, 680
of Ferris Wheel rider, 633
of Great Pyramid of Cheops, 38, 691
of helicopter, 715
of hot-air balloon, 540
of Lincoln’s caricature on Mt. Rushmore,
540
of mountain, 686, 689
of Mt. Everest, 29
of One World Trade Center, 541
of statue on a building, 537
of tower, 540, 541
of tree, 689
of Washington Monument, 540
of Willis Tower, 680
from home, 222
from Honolulu to Melbourne, Australia,
619
of hot-air balloon
to airport, 716
from intersection, 156
from intersection, 264
length
of guy wire, 540, 542, 697
of lake, 602
of ski lift, 689
limiting magnitude of telescope, 500
to the Moon, 690
nautical miles, 518
pendulum swings, 961, 965
to plateau, 540
across a pond, 540
range of airplane, 143
reach of ladder, 540
of rotating beacon, 586
between runners, 689
at sea, 690
of search and rescue, 146
to shore, 540, 602, 690
between skyscrapers, 680
sound to measure, 118–119
stopping, 213, 301, 422
of storm, 145
to tower, 691
traveled by wheel, 37
between two moving vehicles, 156
toward intersection, 264
between two objects, 540
between two planes, 262
visibility of Gibb’s Hill Lighthouse beam,
38, 677–678, 681
visual, 38
walking, 222
width
of gorge, 539
of Mississippi River, 680
of river, 536, 602
Economics
Consumer Price Index (CPI), 477
demand equations, 400
federal stimulus package of 2009, 476
inflation, 476
IS-LM model in, 857
marginal propensity to consume, 966
multiplier, 966
national debt, 235
participation rate, 213
per capita federal debt, 476
poverty rates, 341
poverty threshold, 157
relative income of child, 900
unemployment, 1009
Education
age distribution of community college,
1009
college costs, 476, 965–966
college tuition and fees, 899
computing grades, 129
degrees awarded, 985
doctorates, 1006
faculty composition, 1007
field trip, 374
fraternity purchase, 103
funding a college education, 500
GPA and work relationship, 103
grades, 91
learning curve, 438, 451
maximum level achieved, 937–938
median earnings and level of, 103
multiple-choice test, 995
probability of acceptance to college, 1009
spring break, 931
student loan, 270
interest on, 899
true/false test, 995
video games and grade-point average,
289
Electricity, 91
alternating current (ac), 602, 654
alternating current (ac) circuits, 577, 595
alternating current (ac) generators,
577–578
charging a capacitor, 713
cost of, 243
current in RC circuit, 438
current in RL circuit, 438, 451
impedance, 112
Kirchhoff’s Rules, 857, 873
Ohm’s law, 126
parallel circuits, 112
resistance in, 352
rates for, 129, 180
resistance, 70, 72, 193, 196, 352
voltage
foreign, 29
household, 133
U.S., 29
Electronics. See also Computers and
computing
Blu-ray drive, 516
DVD drive, 516
loudspeakers, 712
microphones, 166
sawtooth curve, 664, 713
Energy
nuclear power plant, 810–811
solar, 166, 771–772
solar heat, 788
thermostat control, 259
Engineering
bridges
clearance, 578
Golden Gate, 305–306
parabolic arch, 317, 788
semielliptical arch, 797, 798, 840
suspension, 309, 787–788
crushing load, 119
drive wheels of engine, 681
electrical, 529
Gateway Arch (St. Louis), 788
grade
of mountain trail, 917
of road, 182
horsepower, 193
lean of Leaning Tower of Pisa, 690
maximum weight supportable by pine,
190
moment of inertia, 668
piston engines, 539
product of inertia, 664
road system, 728
rods and pistons, 698
safe load for a beam, 193
Applications Index
searchlight, 642, 788, 840
whispering galleries, 797
Entertainment
Demon Roller Coaster customer rate,
438
movie theater, 618
theater revenues, 857
Environment
endangered species, 437
lake pollution control laws, 948
oil leakage, 409
Finance, 91. See also Investment(s)
balancing a checkbook, 29
bills in wallet, 1009
cable rates, 494
clothes shopping, 937
college costs, 476, 965–966
computer system purchase, 475
cost
of car, 92, 180
of car rental, 246
of electricity, 243
of fast food, 856
of land, 715
minimizing, 317, 366
of natural gas, 245
of pizza, 92
of printing, 337
of trans-Atlantic travel, 212–213, 221
of triangular lot, 702
cost equation, 192
cost function, 282
cost minimization, 301
credit cards
balance on, 909
debt, 948
interest on, 475
payment, 246, 948
depreciation, 437
of car, 467, 503
discounts, 410
division of money, 88, 91
effective rate of interest, 472
electricity rates, 180
federal stimulus package of 2009, 476
financial planning, 136–137, 856, 869–870,
872
foreign exchange, 410
fraternity purchase, 103
funding a college education, 500
future value of money, 341–342
gross salary, 212
income versus crime rate, 496
inheritance, 146
life cycle hypothesis, 310
loans, 141
car, 948
interest on, 81, 136, 145, 147–148, 270, 899
repayment of, 475
student, 899
median earnings and level of education, 103
mortgages, 477
fees, 246
interest rates on, 476
payments, 189, 192, 196
second, 476
price appreciation of homes, 475
prices of fast food, 858
refunds, 856
revenue equation, 192
revenue maximization, 301, 302–304, 308
rich man’s promise, 966
salary options, 966
sales commission, 128–129
saving
for a car, 475
for a home, 965
savings accounts interest, 475
selling price of a home, 197
sinking fund, 965–966
taxes, 282
e-filing returns, 235
federal income, 246, 410, 422
luxury, 282
truck rentals, 281
used-car purchase, 475
water bills, 129
Food and nutrition
animal, 932
calories, 92
candy, 288
color mix of candy, 1009
cooler contents, 1009
cooling time of pizza, 486
fast food, 856, 858
Girl Scout cookies, 1006
hospital diet, 857, 872
ice cream, 931
“light” foods, 129
number of possible meals, 985–986
raisins, 288–289
soda and hot dogs buying combinations, 283
warming time of beer stein, 487
Forestry
wood product classification, 485
Games
coin toss, 999
die rolling, 998–999, 1000, 1009
grains of wheat on a chess board, 966
lottery, 1009, 1010–1011
Gardens and gardening. See also
Landscaping
border around, 103
enclosure for, 142
Geography
area of Bermuda Triangle, 703
area of lake, 703, 715
inclination of mountain trail, 677, 715
Geology
earthquakes, 452
Geometry
angle between two lines, 654
balloon volume, 409
box volume, 773
circle
area of, 141, 703
center of, 188
circumference of, 28, 141
equation of, 883
inscribed in square, 263
length of chord of, 698
radius of, 188, 916
collinear points, 883
cone volume, 193, 410
cube
length of edge of, 389
surface area of, 29
volume of, 29
cylinder
inscribing in cone, 264
inscribing in sphere, 264
volume of, 193, 410
Descartes’s method of equal roots,
916–917
equation of line, 883
ladder angle, 716
polygon
area of, 883
diagonals of, 103
Pythagorean Theorem, 102
quadrilateral area, 717
rectangle
area of, 28, 212, 261–262, 269
dimensions of, 92, 102, 146, 916
inscribed in circle, 263
inscribed in ellipse, 798
inscribed in semicircle, 263, 664
perimeter of, 28
semicircle inscribed in, 264
semicircle area, 702, 703, 717
sphere
surface area of, 28
volume of, 28
square
area of, 37, 141
diagonals of, 156
perimeter of, 141
surface area
of balloon, 409
of cube, 29
of sphere, 28
triangle
area of, 28, 702, 703, 717, 883
circumscribing, 692
equilateral, 28, 156
inscribed in circle, 264
isosceles, 212, 530, 717, 916
medians of, 155
Pascal’s, 948
perimeter of, 28
xxix
xxx
Applications Index
triangle (continued )
right, 539, 679
sides of, 717
Government
federal debt, 235
per capita, 476
federal income tax, 213, 246, 410, 422
e-filing returns, 235
federal stimulus package of 2009, 476
federal tax withholding, 129
first-class mail, 247
Health. See also Medicine
Landscaping. See also Gardens and
gardening
height of tree, 689
pond enclosure, 317
rectangular pond border, 317
removing stump, 765
tree planting, 872
watering lawn, 516
Law and law enforcement
income vs. crime rate, 496
motor vehicle thefts, 1006
violent crimes, 213
volume of balloon, 409
wire enclosure area, 264
Mixtures. See also Chemistry
blending coffees, 137–138, 141, 147, 925,
935
blending teas, 141
cement, 143
mixed nuts, 141, 856, 925, 935
mixing candy, 141
solutions, 856
water and antifreeze, 142
Motion, 713. See also Physics
age versus total cholesterol, 495
blood pressure, 633
cigarette use among teens, 181
exercising, 129
expenditures on, 213
heartbeats during exercise, 276–277
ideal body weight, 422
life cycle hypothesis, 310
life expectancy, 128
Leisure and recreation
Home improvement. See also
Construction
Measurement
catching a train, 840
on a circle, 516
of Ferris Wheel rider, 633
of golf ball, 220–221
minute hand of clock, 516, 602
objects approaching intersection, 836–837
of pendulum, 713
revolutions of circular disk, 37
simulating, 831
tortoise and the hare race, 916
uniform, 138–139, 141, 836–837
painting a house, 858
optical methods of, 642
of rainfall, 772
Motor vehicles
Housing
Mechanics, 91. See also Physics
apartment rental, 310
number of rooms in, 212
price appreciation of homes, 475
Investment(s), 88, 91, 141, 145
annuity, 962–963, 965
in bonds, 932
Treasuries, 872, 873, 922, 924, 926
zero-coupon, 473, 476
in CDs, 472, 932
compound interest on, 468–469, 470,
471–472, 475–476
diversified, 858
dividing, 247
doubling of, 473–474, 476
effective rate of interest, 472
finance charges, 475
in fixed-income securities, 476, 932
401K, 965, 979
growth rate for, 475–476
IRA, 476, 962–963, 965
mutual fund growth over time, 490
return on, 475, 931, 932
savings account, 471–472
in stock
analyzing, 320
appreciation, 475
beta, 273, 320
NASDAQ stocks, 995
NYSE stocks, 995
portfolios of, 988
price of, 966
time to reach goal, 475, 477
tripling of, 474, 476
amusement park ride, 516
cable TV, 265
rates, 494
community skating rink, 270
Ferris wheel, 187, 633, 691, 712
field trip, 374
video games and grade-point average,
289
Medicine. See also Health
age versus total cholesterol, 495
blood pressure, 633
cancer
breast, 493
pancreatic, 437
drug concentration, 234, 366
drug medication, 437, 451
healing of wounds, 437, 451
spreading of disease, 501
Meteorology
weather balloon height and atmospheric
pressure, 491
Miscellaneous
banquet seating, 931
bending wire, 916
biorhythms, 578
carrying a ladder around a corner, 529,
586, 633–634
citrus ladders, 955
coffee container, 504
cross-sectional area of beam, 213, 220
curve fitting, 857, 872, 935
diameter of copper wire, 29
drafting error, 156
land dimensions, 689
Mandelbrot sets, 751
motor, 29
pet ownership, 1006
reading books, 133
surface area of balloon, 409
alcohol and driving, 447, 452
angular speed of race car, 602
approaching intersection, 836–837
automobile production, 409, 872
average car speed, 143
brake repair with tune-up, 1009
braking load, 772, 775
crankshafts, 690
depreciation, 402
depreciation of, 467, 503
with Global Positioning System (GPS), 501
loans for, 948
markup of new car, 129
runaway car, 315
speed and miles per gallon, 310–311
spin balancing tires, 517
stopping distance, 213, 301, 422
theft of, 1006
used-car purchase, 475
windshield wiper, 516
Music
revenues from, 259
Navigation
avoiding a tropical storm, 697
bearing, 678, 696
of aircraft, 680
of ship, 680, 715
charting a course, 764
commercial, 689
compass heading, 764
crossing a river, 763, 764
error in
correcting, 694–695, 715
time lost due to, 689
Applications Index
rescue at sea, 686–687, 689–690
revising a flight plan, 697
Oceanography
tides, 596
Optics
angle of refraction, 634–635
bending light, 635
Brewster angle, 635
index of refraction, 634–635
intensity of light, 193
laser beam, 679
laser projection, 664
lensmaker’s equation, 72
light obliterated through glass, 437
mirrors, 811
reflecting telescope, 788
Pediatrics
height vs. head circumference, 289,
422
Pharmacy
vitamin intake, 857, 873
Photography
camera distance, 541
Physics, 91
angle of elevation of Sun, 679
angle of inclination, 772
bouncing balls, 979
braking load, 772
damped motion, 708, 716
diameter of atom, 29
Doppler effect, 366
effect of elevation on weight, 221
falling objects, 192
force, 141, 763
of attraction between two bodies, 192
to hold a wagon on a hill, 769–770
muscle, 764
resultant, 763
of wind on a window, 191, 193
gravity, 352, 374
on Earth, 212, 422
on Jupiter, 212
harmonic motion, 707
heat loss, 190, 196
heat transfer, 633
horsepower, 193
inclination of mountain trail, 677
inclined ramp, 764
intensity of light, 146, 193
kinetic energy, 141, 193
maximum weight supportable by pine,
190
missile trajectory, 320
moment of inertia, 668
motion of object, 707
Newton’s law, 192
Ohm’s law, 126
pendulum motion, 119, 516, 713, 961
period, 259–260, 422
simple pendulum, 192
pressure, 141, 192
product of inertia, 664
projectile motion, 102–103, 304–305,
308–309, 539, 552, 633, 634, 659, 664,
668, 759, 829–830, 836, 837, 840
artillery, 315, 624
hit object, 836
thrown object, 836
safe load for a beam, 193
simple harmonic motion, 716
simulating motion, 831
sound to measure distance, 118–119
speed of sound, 133
static equilibrium, 760–761, 764, 765, 775
static friction, 764
stress of materials, 193
stretching a spring, 192
tension, 760–761, 764, 775, 776, 971
thrown object, 146, 759
ball, 310, 314
truck pulls, 765
uniform motion, 141, 146, 264, 836–837,
840
velocity down inclined planes, 80
vertically propelled object, 314
vibrating string, 192
wavelength of visible light, 29
weight, 193, 196
of a boat, 763
of a car, 763
of a piano, 760
work, 141
Psychometrics
IQ tests, 129
Pyrotechnics
fireworks display, 810
Rate. See also Speed
of car, 517
catching a bus, 836
catching a train, 836
current of stream, 857
of emptying
fuel tanks, 146
oil tankers, 143
a pool, 143
a tub, 143
to keep up with the Sun, 518
revolutions per minute
of bicycle wheels, 517
of pulleys, 519
speed
average, 143
of current, 141
of cyclist, 143
of motorboat, 141
of moving walkways, 141–142
per gallon rate and, 310–311
of plane, 143
of sound, 133
of water use, 260
Real estate
Play
commission, 128–129
cost of land, 715
cost of triangular lot, 702
housing prices, 398
mortgage loans, 477
swinging, 717
wagon pulling, 763, 770
Recreation
Population. See also Demographics
bacteria, 439, 486, 493
decline in, 486
E-coli growth, 235, 275
of endangered species, 487
of fruit fly, 484
as function of age, 212
growth in, 486
insect, 352, 486
of trout, 948
of United States, 467, 494, 981
of world, 467, 495, 500, 939
Probability
of birthday shared by people in a room,
487
checkout lines, 1006
classroom composition, 1006
exponential, 433, 437–438, 451
household annual income, 1006
Poisson, 438
“Price is Right” games, 1006
of winning a lottery, 1007
xxxi
bungee jumping, 374
Demon Roller Coaster customer rate, 438
online gambling, 1006
Security
security cameras, 680
Seismology
calibrating instruments, 840
Sequences. See also Combinatorics
ceramic tile floor design, 953–954
Drury Lane Theater, 955
football stadium seating, 955
seats in amphitheater, 955
Speed
of aircraft, 763
angular, 517, 602
of current, 517, 935
as function of time, 222, 264
linear, 514
on Earth, 517
of Moon, 517
xxxii
Applications Index
revolutions per minute of pulley,
517
of rotation of lighthouse beacons,
602
of swimmer, 775
of truck, 679
of wheel pulling cable cars, 517
wind, 856
Sports
baseball, 836, 837, 996, 1008
diamond, 156
dimensions of home plate, 702
field, 697, 698
Little League, 156, 518–519
on-base percentage, 284–285
stadium, 697
World Series, 996
basketball, 996
free throws, 220, 680–681
granny shots, 220
biathlon, 143
bungee jumping, 374
calculating pool shots, 541
distance between runners, 689
exacta betting, 1009
football, 142, 798
defensive squad, 996
field design, 103
seating revenue, 966
golf, 220–221, 496, 829–830, 836
distance to the green, 696
sand bunkers, 624
hammer throw, 603
Olympic heroes, 143
races, 142, 146, 913–914, 916
relay runners, 1008
swimming, 717, 775
tennis, 142
Statistics. See Probability
Surveys
Travel. See also Air travel;
Navigation
of appliance purchases, 987
data analysis, 984, 987
stock portfolios, 988
of summer session attendance, 987
of TV sets in a house, 1006
bearing, 715
drivers stopped by the police, 503
driving to school, 192
parking at O’Hare International
Airport, 245
Temperature
Volume
of air parcel, 955
body, 29, 133
conversion of, 410, 422
cooling time of pizza, 486
cricket chirp rate and, 311
Fahrenheit from Celsius conversion, 87
measuring, 180–181
after midnight, 341
monthly, 595–596, 602
relationship between scales, 259
sinusoidal function from, 591–592
of skillet, 500
warming time of beer stein, 487
wind chill factor, 501
of gasoline in tank, 80
of ice in skating rink, 270
of water in cone, 265
Time
for beer stein to warm, 487
for block to slide down inclined plane, 539
Ferris Wheel rider height as function of, 633
to go from an island to a town, 265
hours of daylight, 400–401, 505, 593–594,
597, 604, 618
for pizza to cool, 486
for rescue at sea, 146
of sunrise, 518, 618
of trip, 529, 541
Transportation
deicing salt, 624
Niagara Falls Incline Railway, 680
Weapons
artillery, 315, 624
cannons, 320
Weather
atmospheric pressure, 437,
451
avoiding a tropical storm, 697
cooling air, 955
hurricanes, 341, 595
lightning and thunder, 145
lightning strikes, 807–808, 810
probability of rain, 1002
rainfall measurement, 772
relative humidity, 438
weather satellites, 187
wind chill, 246–247, 501
Work, 770
computing, 770, 771, 775
constant rate jobs, 936
GPA and, 103
pulling a wagon, 770
ramp angle, 772
wheelbarrow push, 763
working together, 140, 142, 146
R
Review
A Look Ahead
Outline
Chapter R, as the title states, contains review material. Your instructor may
choose to cover all or part of it as a regular chapter at the beginning of your
course or later as a just-in-time review when the content is required. Regardless,
when information in this chapter is needed, a specific reference to this chapter
will be made so you can review.
R.1
R.2
R.3
R.4
R.5
R.6
R.7
R.8
Real Numbers
Algebra Essentials
Geometry Essentials
Polynomials
Factoring Polynomials
Synthetic Division
Rational Expressions
nth Roots; Rational Exponents
1
1
2
CHAPTER R Review
R.1 Real Numbers
PREPARING FOR THIS TEXT Before getting started, read “To the Student” at the front of this
text.
OBJECTIVES 1
2
3
4
Work with Sets (p. 2)
Classify Numbers (p. 4)
Evaluate Numerical Expressions (p. 8)
Work with Properties of Real Numbers (p. 9)
1 Work with Sets
A set is a well-defined collection of distinct objects. The objects of a set are called its
elements. By well-defined, we mean that there is a rule that enables us to determine
whether a given object is an element of the set. If a set has no elements, it is called
the empty set, or null set, and is denoted by the symbol ∅.
For example, the set of digits consists of the collection of numbers 0, 1, 2, 3, 4,
5, 6, 7, 8, and 9. If we use the symbol D to denote the set of digits, then we can write
D = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96
In this notation, the braces 5 6 are used to enclose the objects, or elements, in
the set. This method of denoting a set is called the roster method. A second way to
denote a set is to use set-builder notation, where the set D of digits is written as
D =
{
x
0 x is a digit }
c
c
c
6
$%+++&
Read as “D is the set of all x such that x is a digit.”
2
EX AMPLE 1
Using Set-builder Notation and the Roster Method
(a) E = 5 x 0 x is an even digit6 = 5 0, 2, 4, 6, 86
(b) O = 5 x 0 x is an odd digit6 = 5 1, 3, 5, 7, 96
r
Because the elements of a set are distinct, we never repeat elements. For
example, we would never write 5 1, 2, 3, 26 ; the correct listing is 5 1, 2, 36 . Because
a set is a collection, the order in which the elements are listed is immaterial. 5 1, 2, 36 ,
5 1, 3, 26 , 5 2, 1, 36 , and so on, all represent the same set.
If every element of a set A is also an element of a set B, then A is a subset of B,
which is denoted A ⊆ B. If two sets A and B have the same elements, then A equals
B, which is denoted A = B.
For example, 5 1, 2, 36 ⊆ 5 1, 2, 3, 4, 56 and 5 1, 2, 36 = 5 2, 3, 16 .
DEFINITION
EX AMPLE 2
If A and B are sets, the intersection of A with B, denoted A ∩ B, is the set
consisting of elements that belong to both A and B. The union of A with B,
denoted A ∪ B, is the set consisting of elements that belong to either A or B,
or both.
Finding the Intersection and Union of Sets
Let A = 5 1, 3, 5, 86 , B = 5 3, 5, 76 , and C = 5 2, 4, 6, 86 . Find:
(a) A ∩ B
(b) A ∪ B
(c) B ∩ 1A ∪ C2
SECTION R.1 Real Numbers
Solution
(a) A ∩ B = 5 1, 3, 5, 86 ∩ 5 3, 5, 76 = 5 3, 56
(b) A ∪ B = 5 1, 3, 5, 86 ∪ 5 3, 5, 76 = 5 1, 3, 5, 7, 86
(c) B ∩ 1A ∪ C2 = 5 3, 5, 76 ∩ 3 5 1, 3, 5, 86 ∪ 5 2, 4, 6, 86 4
= 5 3, 5, 76 ∩ 5 1, 2, 3, 4, 5, 6, 86 = 5 3, 56
Now Work
PROBLEM
3
r
15
Usually, in working with sets, we designate a universal set U, the set consisting of
all the elements that we wish to consider. Once a universal set has been designated,
we can consider elements of the universal set not found in a given set.
DEFINITION
EXAMPL E 3
If A is a set, the complement of A, denoted A, is the set consisting of all the
elements in the universal set that are not in A.*
Finding the Complement of a Set
If the universal set is U = 5 1, 2, 3, 4, 5, 6, 7, 8, 96 and if A = 5 1, 3, 5, 7, 96 , then
A = 5 2, 4, 6, 86 .
r
It follows from the definition of complement that A ∪ A = U and A ∩ A = ∅.
Do you see why?
Now Work
Universal set
B
A
PROBLEM
19
It is often helpful to draw pictures of sets. Such pictures, called Venn diagrams,
represent sets as circles enclosed in a rectangle, which represents the universal
set. Such diagrams often help us to visualize various relationships among sets. See
Figure 1.
If we know that A ⊆ B, we might use the Venn diagram in Figure 2(a). If we
know that A and B have no elements in common—that is, if A ∩ B = ∅—we might
use the Venn diagram in Figure 2(b). The sets A and B in Figure 2(b) are said to be
disjoint.
C
Figure 1 Venn diagram
Universal set
Universal set
B
A
A
(a) A 債 B
subset
Figure 2
B
(b) A 傽 B 5 ⭋
disjoint sets
Figures 3(a), 3(b), and 3(c) use Venn diagrams to illustrate the definitions of
intersection, union, and complement, respectively.
Universal set
A
Figure 3
B
(a) A B
intersection
Universal set
Universal set
A
B
(b) A B
union
*Some books use the notation A′ for the complement of A.
A
A
(c) A
complement
4
CHAPTER R Review
2 Classify Numbers
It is helpful to classify the various kinds of numbers that we deal with as sets. The
counting numbers, or natural numbers, are the numbers in the set 5 1, 2, 3, 4, c6 .
(The three dots, called an ellipsis, indicate that the pattern continues indefinitely.)
As their name implies, these numbers are often used to count things. For example,
there are 26 letters in our alphabet; there are 100 cents in a dollar. The whole
numbers are the numbers in the set 5 0, 1, 2, 3, c6 —that is, the counting numbers
together with 0. The set of counting numbers is a subset of the set of whole numbers.
DEFINITION
The integers are the set of numbers 5 c, - 3, - 2, - 1, 0, 1, 2, 3,c6 .
These numbers are useful in many situations. For example, if your checking account
has $10 in it and you write a check for $15, you can represent the current balance
as −$5.
Each time we expand a number system, such as from the whole numbers to the
integers, we do so in order to be able to handle new, and usually more complicated,
problems. The integers enable us to solve problems requiring both positive and
negative counting numbers, such as profit/loss, height above/below sea level,
temperature above/below 0°F, and so on.
But integers alone are not sufficient for all problems. For example, they do not
answer the question “What part of a dollar is 38 cents?” To answer such a question,
38
we enlarge our number system to include rational numbers. For example,
100
answers the question “What part of a dollar is 38 cents?”
DEFINITION
a
of two
b
integers. The integer a is called the numerator, and the integer b, which cannot
be 0, is called the denominator. The rational numbers are the numbers in the
a
set e x ` x = , where a, b are integers and b ≠ 0 f.
b
A rational number is a number that can be expressed as a quotient
3 5 0
2
100
a
, , , - , and
. Since = a for any
4 2 4
3
3
1
integer a, it follows that the set of integers is a subset of the set of rational numbers.
Rational numbers may be represented as decimals. For example, the rational
2
7
3 5
may be represented as decimals by merely carrying out
numbers , , - , and
4 2
3
66
the indicated division:
Examples of rational numbers are
3
= 0.75
4
5
= 2.5
2
-
2
= - 0.666c = - 0.6
3
7
= 0.1060606c = 0.106
66
5
3
Notice that the decimal representations of and terminate, or end. The decimal
4
2
7
2
representations of - and
do not terminate, but they do exhibit a pattern of
66
2 3
repetition. For - , the 6 repeats indefinitely, as indicated by the bar over the 6; for
3
7
, the block 06 repeats indefinitely, as indicated by the bar over the 06. It can be
66
shown that every rational number may be represented by a decimal that either
terminates or is nonterminating with a repeating block of digits, and vice versa.
On the other hand, some decimals do not fit into either of these categories. Such
decimals represent irrational numbers. Every irrational number may be represented
by a decimal that neither repeats nor terminates. In other words, irrational numbers
a
cannot be written in the form , where a, b are integers and b ≠ 0.
b
SECTION R.1 Real Numbers
5
Irrational numbers occur naturally. For example, consider the isosceles right
triangle whose legs are each of length 1. See Figure 4. The length of the hypotenuse
is 12, an irrational number.
Also, the number that equals the ratio of the circumference C to the diameter d
of any circle, denoted by the symbol p (the Greek letter pi), is an irrational number.
See Figure 5.
C
2
1
d
1
Figure 5 p =
Figure 4
DEFINITION
C
d
The set of real numbers is the union of the set of rational numbers with the set
of irrational numbers.
Figure 6 shows the relationship of various types of numbers.*
Irrational numbers
Rational
numbers
Integers
Whole numbers
Natural or
counting
numbers
Real numbers
Figure 6
EXAMPL E 4
Classifying the Numbers in a Set
List the numbers in the set
4
b - 3, , 0.12, 22, p, 10, 2.151515c1where the block 15 repeats2 r
3
that are
(a) Natural numbers
(d) Irrational numbers
Solution
(b) Integers
(e) Real numbers
(c) Rational numbers
(a) 10 is the only natural number.
(b) - 3 and 10 are integers.
4
(c) - 3, 10, , 0.12, and 2.151515c are rational numbers.
3
(d) 12 and p are irrational numbers.
(e) All the numbers listed are real numbers.
Now Work
PROBLEM
25
*The set of real numbers is a subset of the set of complex numbers. We discuss complex numbers in
Chapter 1, Section 1.3.
r
6
CHAPTER R Review
Approximations
Every decimal may be represented by a real number (either rational or irrational),
and every real number may be represented by a decimal.
In practice, the decimal representation of an irrational number is given as an
approximation. For example, using the symbol ≈ (read as “approximately equal
to”), we can write
22 ≈ 1.4142
p ≈ 3.1416
In approximating decimals, we either round off or truncate to a given number of
decimal places.* The number of places establishes the location of the final digit in
the decimal approximation.
Truncation: Drop all of the digits that follow the specified final digit in the
decimal.
Rounding: Identify the specified final digit in the decimal. If the next digit is
5 or more, add 1 to the final digit; if the next digit is 4 or less, leave the final digit
as it is. Then truncate following the final digit.
EX AMPLE 5
Approximating a Decimal to Two Places
Approximate 20.98752 to two decimal places by
(a) Truncating
(b) Rounding
Solution
For 20.98752, the final digit is 8, since it is two decimal places from the decimal point.
(a) To truncate, we remove all digits following the final digit 8. The truncation of
20.98752 to two decimal places is 20.98.
(b) The digit following the final digit 8 is the digit 7. Since 7 is 5 or more, we add
1 to the final digit 8 and truncate. The rounded form of 20.98752 to two decimal
places is 20.99.
r
EX AMPLE 6
Approximating a Decimal to Two and Four Places
Rounded
to Two
Decimal
Places
Number
Rounded
to Four
Decimal
Places
Truncated
to Two
Decimal
Places
Truncated
to Four
Decimal
Places
(a)
3.14159
3.14
3.1416
3.14
3.1415
(b)
0.056128
0.06
0.0561
0.05
0.0561
893.46
893.4613
893.46
893.4612
(c) 893.46125
Now Work
PROBLEM
r
29
Calculators
Calculators are incapable of displaying decimals that contain a large number of digits.
For example, some calculators are capable of displaying only eight digits. When a
number requires more than eight digits, the calculator either truncates or rounds.
* Sometimes we say “correct to a given number of decimal places” instead of “truncate.”
SECTION R.1 Real Numbers
7
To see how your calculator handles decimals, divide 2 by 3. How many digits do you
see? Is the last digit a 6 or a 7? If it is a 6, your calculator truncates; if it is a 7, your
calculator rounds.
There are different kinds of calculators. An arithmetic calculator can only add,
subtract, multiply, and divide numbers; therefore, this type is not adequate for this
course. Scientific calculators have all the capabilities of arithmetic calculators and
also contain function keys labeled ln, log, sin, cos, tan, xy, inv, and so on. As you
proceed through this text, you will discover how to use many of the function keys.
Graphing calculators have all the capabilities of scientific calculators and contain a
screen on which graphs can be displayed.
For those who have access to a graphing calculator, we have included comments,
examples, and exercises marked with a , indicating that a graphing calculator is
required. We have also included an appendix that explains some of the capabilities
of a graphing calculator. The comments, examples, and exercises may be omitted
without loss of continuity, if so desired.
Operations
In algebra, we use letters such as x, y, a, b, and c to represent numbers. The symbols
used in algebra for the operations of addition, subtraction, multiplication, and division
are +, - , # , and >. The words used to describe the results of these operations are
sum, difference, product, and quotient. Table 1 summarizes these ideas.
Table 1
Operation
Symbol
Words
Addition
a + b
Sum: a plus b
Subtraction
a - b
Difference: a minus b
Multiplication
a # b, (a) # b, a # (b), (a) # (b),
ab, (a)b, a(b), (a)(b)
Product: a times b
Division
a>b or
a
b
Quotient: a divided by b
In algebra, we generally avoid using the multiplication sign * and the division
sign , so familiar in arithmetic. Notice also that when two expressions are placed
next to each other without an operation symbol, as in ab, or in parentheses, as in
1a2 1b2, it is understood that the expressions, called factors, are to be multiplied.
We also prefer not to use mixed numbers in algebra. When mixed numbers
3
3
are used, addition is understood; for example, 2 means 2 + . In algebra, use of
4
4
a mixed number may be confusing because the absence of an operation symbol
3
between two terms is generally taken to mean multiplication. The expression 2 is
4
11
therefore written instead as 2.75 or as .
4
The symbol =, called an equal sign and read as “equals” or “is,” is used to
express the idea that the number or expression on the left of the equal sign is
equivalent to the number or expression on the right.
EXAMPL E 7
Writing Statements Using Symbols
(a) The sum of 2 and 7 equals 9. In symbols, this statement is written as 2 + 7 = 9.
(b) The product of 3 and 5 is 15. In symbols, this statement is written as 3 # 5 = 15.
Now Work
PROBLEM
41
r
8
CHAPTER R Review
3 Evaluate Numerical Expressions
Consider the expression 2 + 3 # 6. It is not clear whether we should add 2 and 3 to
get 5, and then multiply by 6 to get 30; or first multiply 3 and 6 to get 18, and then
add 2 to get 20. To avoid this ambiguity, we have the following agreement.
In Words
Multiply first, then add.
We agree that whenever the two operations of addition and multiplication
separate three numbers, the multiplication operation will always be performed
first, followed by the addition operation.
For 2 + 3 # 6, then, we have
2 + 3 # 6 = 2 + 18 = 20
EX AMPLE 8
Finding the Value of an Expression
Evaluate each expression.
(a) 3 + 4 # 5
Solution
(b) 8 # 2 + 1
(a) 3 + 4 # 5 = 3 + 20 = 23
(c) 2 + 2 # 2
(b) 8 # 2 + 1 = 16 + 1 = 17
c
Multiply first
c
Multiply first
r
(c) 2 + 2 # 2 = 2 + 4 = 6
Now Work
PROBLEM
53
When we want to indicate adding 3 and 4 and then multiplying the result by 5,
we use parentheses and write 13 + 42 # 5. Whenever parentheses appear in an
expression, it means “perform the operations within the parentheses first!”
EX AMPLE 9
Finding the Value of an Expression
(a) 15 + 32 # 4 = 8 # 4 = 32
(b) 14 + 52 # 18 - 22 = 9 # 6 = 54
r
When we divide two expressions, as in
2 + 3
4 + 8
it is understood that the division bar acts like parentheses; that is,
12 + 32
2 + 3
=
4 + 8
14 + 82
Rules for the Order of Operations
1. Begin with the innermost parentheses and work outward. Remember that
in dividing two expressions, we treat the numerator and denominator as if
they were enclosed in parentheses.
2. Perform multiplications and divisions, working from left to right.
3. Perform additions and subtractions, working from left to right.
SECTION R.1 Real Numbers
EX AM PL E 1 0
Finding the Value of an Expression
Evaluate each expression.
(b) 5 # 13 + 42 + 2
(a) 8 # 2 + 3
2 + 5
(c)
2 + 4#7
Solution
9
(d) 2 + 3 4 + 2 # 110 + 62 4
(a) 8 # 2 + 3 = 16 + 3 = 19
c
Multiply first
(b) 5 # 13 + 42 + 2 = 5 # 7 + 2 = 35 + 2 = 37
c
c
Parentheses first
Multiply before adding
2 + 5
2 + 5
7
=
=
#
2 + 4 7
2 + 28
30
(d) 2 + 3 4 + 2 # 110 + 62 4 = 2 + 3 4 + 2 # 1162 4
= 2 + 3 4 + 324 = 2 + 3 364 = 38
(c)
r
Be careful if you use a calculator. For Example 10(c), you need to use parentheses.
See Figure 7.* If you don’t, the calculator will compute the expression
2 +
5
+ 4 # 7 = 2 + 2.5 + 28 = 32.5
2
giving a wrong answer.
Now Work
Figure 7
PROBLEMS
59
AND
67
4 Work with Properties of Real Numbers
The equal sign is used to mean that one expression is equivalent to another. Four
important properties of equality are listed next. In this list, a, b, and c represent real
numbers.
1.
2.
3.
4.
The reflexive property states that a number equals itself; that is, a = a.
The symmetric property states that if a = b, then b = a.
The transitive property states that if a = b and b = c, then a = c.
The principle of substitution states that if a = b, then we may substitute b
for a in any expression containing a.
Now, let’s consider some other properties of real numbers.
EXAMPL E 11
Commutative Properties
(a) 3 + 5 = 8
5 + 3 = 8
3 + 5 = 5 + 3
(b) 2 # 3 = 6
3#2 = 6
2#3 = 3#2
r
This example illustrates the commutative property of real numbers, which states
that the order in which addition or multiplication takes place does not affect the
final result.
* Notice that we converted the decimal to its fraction form. Another option, when using a TI-84 Plus C,
is to use the fraction template under the MATH button to enter the expression as it appears in Example 10(c).
Consult your manual to see how to enter such expressions on your calculator.
10
CHAPTER R Review
Commutative Properties
a + b = b + a
(1a)
a#b = b#a
(1b)
Here, and in the properties listed next and on pages 11–13, a, b, and c represent
real numbers.
EX A MPL E 1 2
Associative Properties
(a) 2 + 13 + 42 = 2 + 7 = 9
12 + 32 + 4 = 5 + 4 = 9
2 + 13 + 42 = 12 + 32 + 4
(b) 2 # 13 # 42 = 2 # 12 = 24
12 # 32 # 4 = 6 # 4 = 24
2 # 13 # 42 = 12 # 32 # 4
r
The way we add or multiply three real numbers does not affect the final result.
Expressions such as 2 + 3 + 4 and 3 # 4 # 5 present no ambiguity, even though addition
and multiplication are performed on one pair of numbers at a time. This property is
called the associative property.
Associative Properties
a + 1b + c2 = 1a + b2 + c = a + b + c
a # 1b # c2 = 1a # b2 # c = a # b # c
(2a)
(2b)
Distributive Property
a # 1b + c2 = a # b + a # c
1a + b2 # c = a # c + b # c
(3a)
(3b)
The distributive property may be used in two different ways.
EX A MPL E 1 3
Distributive Property
(a) 2 # 1x + 32 = 2 # x + 2 # 3 = 2x + 6 Use to remove parentheses.
Use to combine two expressions.
(b) 3x + 5x = 13 + 52x = 8x
(c) 1x + 22 1x + 32 = x1x + 32 + 21x + 32 = 1x2 + 3x2 + 12x + 62
= x2 + (3x + 2x) + 6 = x2 + 5x + 6
Now Work
PROBLEM
r
89
The real numbers 0 and 1 have unique properties called the identity properties.
EX A MPL E 1 4
Identity Properties
(a) 4 + 0 = 0 + 4 = 4
(b) 3 # 1 = 1 # 3 = 3
r
SECTION R.1 Real Numbers
11
Identity Properties
0 + a = a + 0 = a
(4a)
a#1 = 1#a = a
(4b)
We call 0 the additive identity and 1 the multiplicative identity.
For each real number a, there is a real number - a, called the additive inverse
of a, having the following property:
Additive Inverse Property
a + 1 - a2 = - a + a = 0
EX AM PL E 15
(5a)
Finding an Additive Inverse
(a) The additive inverse of 6 is - 6, because 6 + 1 - 62 = 0.
(b) The additive inverse of - 8 is - 1 - 82 = 8, because - 8 + 8 = 0.
r
The additive inverse of a, that is, - a, is often called the negative of a or the
opposite of a. The use of such terms can be dangerous, because they suggest that
the additive inverse is a negative number, which may not be the case. For example, the
additive inverse of - 3, or - 1 - 32, equals 3, a positive number.
1
For each nonzero real number a, there is a real number , called the multiplicative
a
inverse of a, having the following property:
Multiplicative Inverse Property
a#
1
1
= #a = 1
a
a
The multiplicative inverse
if a ≠ 0
(5b)
1
of a nonzero real number a is also referred to as the
a
reciprocal of a.
EX AM PL E 1 6
Finding a Reciprocal
1
1
(a) The reciprocal of 6 is , because 6 # = 1.
6
6
1
1
(b) The reciprocal of - 3 is
, because - 3 #
= 1.
-3
-3
2 3
2 3
(c) The reciprocal of is , because # = 1.
3 2
3 2
r
With these properties for adding and multiplying real numbers, we can define
the operations of subtraction and division as follows:
DEFINITION
The difference a - b, also read “a less b” or “a minus b,” is defined as
a - b = a + 1 - b2
To subtract b from a, add the opposite of b to a.
(6)
12
CHAPTER R Review
DEFINITION
a
If b is a nonzero real number, the quotient , also read as “a divided by b” or
b
“the ratio of a to b,” is defined as
a
1
= a#
b
b
EX A MPL E 1 7
if b ≠ 0
(7)
Working with Differences and Quotients
(a) 8 - 5 = 8 + 1 - 52 = 3
(b) 4 - 9 = 4 + 1 - 92 = - 5
5
1
(c)
= 5#
8
8
r
For any number a, the product of a times 0 is always 0; that is,
In Words
Multiplication by Zero
a#0 = 0
The result of multiplying by zero
is zero.
(8)
For a nonzero number a,
Division Properties
0
= 0
a
a
= 1 if a ≠ 0
a
(9)
2
= x means to
0
find x such that 0 # x = 2. But 0 # x equals 0 for all x, so there is no unique number x such that
2
= x.
■
0
NOTE Division by 0 is not defined. One reason is to avoid the following difficulty:
Rules of Signs
a1 - b2 = - 1ab2
- 1 - a2 = a
EX A MPL E 1 8
Applying the Rules of Signs
(a) 21 - 32 = - 12 # 32 = - 6
3
-3
3
(c)
=
= -2
2
2
x
1 #
1
(e)
=
x = - x
-2
-2
2
1 - a2b = - 1ab2
a
-a
a
=
= -b
b
b
1 - a2 1 - b2 = ab
-a
a
=
(10)
-b
b
(b) 1 - 32 1 - 52 = 3 # 5 = 15
-4
4
(d)
=
-9
9
r
SECTION R.1 Real Numbers
13
Cancellation Properties
ac = bc implies a = b if c ≠ 0
ac
a
=
if b ≠ 0, c ≠ 0
bc
b
EX AM PL E 1 9
(11)
Using the Cancellation Properties
(a) If 2x = 6, then
2x = 6
2x = 2 # 3
x = 3
NOTE We follow the common practice
of using slash marks to indicate
cancellations.
■
In Words
(b)
18
=
12
3# 6
2# 6
Factor 6.
Cancel the 2’s.
3
2
c
=
r
Cancel the 6’s.
Zero-Product Property
If a product equals 0, then one or
both of the factors is 0.
EX AM PL E 2 0
If ab = 0, then a = 0, or b = 0, or both.
(12)
Using the Zero-Product Property
r
If 2x = 0, then either 2 = 0 or x = 0. Since 2 ≠ 0, it follows that x = 0.
Arithmetic of Quotients
a
c
ad
bc
ad + bc
+
=
+
=
b
d
bd
bd
bd
a#c
ac
=
b d
bd
a
a d
b
ad
= # =
c
b c
bc
d
EX AM PL E 2 1
if b ≠ 0, d ≠ 0
(13)
if b ≠ 0, d ≠ 0
(14)
if b ≠ 0, c ≠ 0, d ≠ 0
(15)
Adding, Subtracting, Multiplying, and Dividing Quotients
(a)
2
5
2#2
3#5
2#2 + 3#5
4 + 15
19
+ = # + # =
=
=
#
3
2
3 2
3 2
3 2
6
6
c By equation (13)
(b)
3
2
3
2
3
-2
- =
+ a- b = +
5
3
5
3
5
3
c
By equation (6)
=
c
By equation (10)
3 # 3 + 5 # 1 - 22
9 + 1 - 102
-1
1
=
=
= #
5 3
15
15
15
c By equation (13)
14
CHAPTER R Review
NOTE Slanting the cancellation marks
in different directions for different
factors, as shown here, is a good
practice to follow, since it will help in
checking for errors.
■
(c)
2#5
8 # 15
8 # 15
2# 4 # 3 #5
=
= 10
= #
=
#
#
3 4 1
1
3 4
3 4
c
c
By equation (14)
By equation (11)
3
5
3 9
3#9
27
(d)
= # = # =
7
5 7 c 5 7
35
By
equation
(14)
9 æ
r
By equation (15)
NOTE In writing quotients, we shall follow the usual convention and write the quotient in lowest terms.
That is, we write it so that any common factors of the numerator and the denominator have been
removed using the cancellation properties, equation (11). As examples,
90
15 # 6
15
=
=
24
4# 6
4
4# 6 #x# x
4x
24x 2
=
=
18x
3# 6 # x
3
Now Work
PROBLEMS
69, 73,
AND
x ≠ 0
■
83
Sometimes it is easier to add two fractions using least common multiples (LCM).
The LCM of two numbers is the smallest number that each has as a common
multiple.
EX A MPL E 2 2
Finding the Least Common Multiple of Two Numbers
Find the least common multiple of 15 and 12.
Solution
To find the LCM of 15 and 12, we look at multiples of 15 and 12.
15, 30, 45, 60, 75, 90, 105, 120,c
12, 24, 36, 48, 60, 72, 84, 96, 108, 120,c
The common multiples are in blue. The least common multiple is 60.
EX A MPL E 2 3
Using the Least Common Multiple to Add Two Fractions
Find:
Solution
r
8
5
+
15
12
We use the LCM of the denominators of the fractions and rewrite each fraction using
the LCM as a common denominator. The LCM of the denominators (12 and 15)
is 60. Rewrite each fraction using 60 as the denominator.
8
5
8 #4
5 #5
+
=
+
15
12
15 4
12 5
32
25
=
+
60
60
32 + 25
=
60
57
=
60
19
=
20
Now Work
PROBLEM
77
r
SECTION R.1 Real Numbers
15
Historical Feature
T
he real number system has a history that stretches back at least to
the ancient Babylonians (1800 BC). It is remarkable how much the
ancient Babylonian attitudes resemble our own. As we stated in
the text, the fundamental difficulty with irrational numbers is that they
cannot be written as quotients of integers or, equivalently, as repeating
or terminating decimals. The Babylonians wrote their numbers in a
system based on 60 in the same way that we write ours based on 10.
They would carry as many places for p as the accuracy of the problem
demanded, just as we now use
p ≈ 3
1
7
or p ≈ 3.1416 or p ≈ 3.14159
or p ≈ 3.14159265358979
depending on how accurate we need to be.
Things were very different for the Greeks, whose number system
allowed only rational numbers. When it was discovered that 12 was not
a rational number, this was regarded as a fundamental flaw in the number
concept. So serious was the matter that the Pythagorean Brotherhood
(an early mathematical society) is said to have drowned one of its
members for revealing this terrible secret. Greek mathematicians then
turned away from the number concept, expressing facts about whole
numbers in terms of line segments.
In astronomy, however, Babylonian methods, including the Babylonian
number system, continued to be used. Simon Stevin (1548–1620),
probably using the Babylonian system as a model, invented the decimal
system, complete with rules of calculation, in 1585. [Others, for example,
al-Kashi of Samarkand (d. 1429), had made some progress in the same
direction.] The decimal system so effectively conceals the difficulties
that the need for more logical precision began to be felt only in the early
1800s. Around 1880, Georg Cantor (1845–1918) and Richard Dedekind
(1831–1916) gave precise definitions of real numbers. Cantor’s definition,
although more abstract and precise, has its roots in the decimal (and
hence Babylonian) numerical system.
Sets and set theory were a spin-off of the research that went into
clarifying the foundations of the real number system. Set theory has
developed into a large discipline of its own, and many mathematicians
regard it as the foundation upon which modern mathematics is built.
Cantor’s discoveries that infinite sets can also be counted and that there
are different sizes of infinite sets are among the most astounding results
of modern mathematics.
R.1 Assess Your Understanding
Concepts and Vocabulary
1. The numbers in the set e x ` x =
and b ≠ 0 f are called
a
, where a, b are integers
b
numbers.
2. The value of the expression 4 + 5 # 6 - 3 is
.
3. The fact that 2x + 3x = 12 + 32x is a consequence of the
Property.
4. “The product of 5 and x + 3 equals 6” may be written as
.
5. The intersection of sets A and B is denoted by which of the
following?
(a) A ∩ B (b) A ∪ B (c) A ⊆ B (d) A ∅ B
6. Choose the correct name for the set of numbers
5 0, 1, 2, 3, c6 .
(a) Counting numbers
(b) Whole numbers
(c) Integers
(d) Irrational numbers
7. True or False Rational numbers have decimals that either
terminate or are nonterminating with a repeating block of
digits.
8. True or False The Zero-Product Property states that the
product of any number and zero equals zero.
9. True or False The least common multiple of 12 and 18
is 6.
10. True or False No real number is both rational and
irrational.
Skill Building
In Problems 11–22, use U = universal set = 50, 1, 2, 3, 4, 5, 6, 7, 8, 96, A = 51, 3, 4, 5, 96 , B = 52, 4, 6, 7, 86 , and C = 51, 3, 4, 66
to find each set.
15. 1A ∪ B2 ∩ C
16. 1A ∩ B2 ∪ C
13. A ∩ B
14. A ∩ C
17. A
18. C
19. A ∩ B
20. B ∪ C
21. A ∪ B
22. B ∩ C
11. A ∪ B
12. A ∪ C
In Problems 23–28, list the numbers in each set that are (a) Natural numbers, (b) Integers, (c) Rational numbers, (d) Irrational numbers,
(e) Real numbers.
1
23. A = e - 6, , - 1.333c1the 3>s repeat2, p, 2, 5 f
2
5
24. B = e - , 2.060606c1the block 06 repeats2, 1.25, 0, 1, 25 f
3
1 1 1
25. C = e 0, 1, , , f
2 3 4
26. D = 5 - 1, - 1.1, - 1.2, - 1.36
27. E = e 22, p, 22 + 1, p +
1
f
2
28. F = e - 22, p + 22,
1
+ 10.3 f
2
16
CHAPTER R Review
In Problems 29–40, approximate each number (a) rounded and (b) truncated to three decimal places.
29. 18.9526
30. 25.86134
31. 28.65319
35. 9.9985
36. 1.0006
37.
3
7
32. 99.05249
38.
5
9
33. 0.06291
39.
34. 0.05388
521
15
40.
81
5
In Problems 41–50, write each statement using symbols.
41. The sum of 3 and 2 equals 5.
42. The product of 5 and 2 equals 10.
43. The sum of x and 2 is the product of 3 and 4.
44. The sum of 3 and y is the sum of 2 and 2.
45. The product of 3 and y is the sum of 1 and 2.
46. The product of 2 and x is the product of 4 and 6.
47. The difference x less 2 equals 6.
48. The difference 2 less y equals 6.
49. The quotient x divided by 2 is 6.
50. The quotient 2 divided by x is 6.
In Problems 51–88, evaluate each expression.
54. 8 - 4 # 2
51. 9 - 4 + 2
52. 6 - 4 + 3
53. - 6 + 4 # 3
55. 4 + 5 - 8
56. 8 - 3 - 4
57. 4 +
59. 6 - 33 # 5 + 2 # 13 - 22 4
60. 2 # 38 - 314 + 22 4 - 3
61. 2 # 13 - 52 + 8 # 2 - 1
63. 10 - 36 - 2 # 2 + 18 - 32 4 # 2
65. 15 - 32
58. 2 -
1
2
62. 1 - 14 # 3 - 2 + 22
64. 2 - 5 # 4 - 36 # 13 - 42 4
66. 15 + 42
1
2
1
3
1
3
67.
4 + 8
5 - 3
68.
2 - 4
5 - 3
69.
3 # 10
5 21
70.
5# 3
9 10
71.
6 # 10
25 27
72.
21 # 100
25 3
73.
3
2
+
4
5
74.
4
1
+
3
2
75.
5
9
+
6
5
76.
8
15
+
9
2
77.
5
1
+
18
12
78.
2
8
+
15
9
79.
1
7
30
18
80.
3
2
14
21
3
2
81.
20
15
85.
7
1#3
+
2 5
10
6
3
82.
35
14
5
18
83.
11
27
2
4 1
+ #
3
5 6
87. 2 #
86.
5
21
84.
2
35
3
3
+
4
8
88. 3 #
5
1
6
2
In Problems 89–100, use the Distributive Property to remove the parentheses.
89. 61x + 42
90. 412x - 12
91. x1x - 42
92. 4x1x + 32
1
3
93. 2a x - b
4
2
2
1
94. 3a x + b
3
6
95. 1x + 22 1x + 42
96. 1x + 52 1x + 12
97. 1x - 22 1x + 12
98. 1x - 42 1x + 12
99. 1x - 82 1x - 22
100. 1x - 42 1x - 22
Explaining Concepts: Discussion and Writing
101. Explain to a friend how the Distributive Property is used to
justify the fact that 2x + 3x = 5x.
102. Explain to a friend
12 + 32 # 4 = 20.
why
2 + 3 # 4 = 14,
whereas
103. Explain why 213 # 42 is not equal to 12 # 32 # 12 # 42.
104. Explain why
4 + 3
4
3
is not equal to + .
2 + 5
2
5
SECTION R.2 Algebra Essentials
105. Is subtraction commutative? Support your conclusion with
an example.
106. Is subtraction associative? Support your conclusion with an
example.
107. Is division commutative? Support your conclusion with an
example.
108. Is division associative? Support your conclusion with an
example.
109. If 2 = x, why does x = 2?
17
113. A rational number is defined as the quotient of two integers.
When written as a decimal, the decimal will either repeat
or terminate. By looking at the denominator of the rational
number, there is a way to tell in advance whether its decimal
representation will repeat or terminate. Make a list of
rational numbers and their decimals. See if you can discover
the pattern. Confirm your conclusion by consulting books
on number theory at the library. Write a brief essay on your
findings.
114. The current time is 12 noon CST. What time (CST) will it be
12,997 hours from now?
110. If x = 5, why does x2 + x = 30?
111. Are there any real numbers that are both rational and
irrational? Are there any real numbers that are neither?
Explain your reasoning.
0
a
1a ≠ 02 and
are undefined, but for different
0
0
reasons. Write a paragraph or two explaining the different
reasons.
115. Both
112. Explain why the sum of a rational number and an irrational
number must be irrational.
R.2 Algebra Essentials
OBJECTIVES 1 Graph Inequalities (p. 18)
2 Find Distance on the Real Number Line (p. 19)
3 Evaluate Algebraic Expressions (p. 20)
4 Determine the Domain of a Variable (p. 21)
5 Use the Laws of Exponents (p. 21)
6 Evaluate Square Roots (p. 23)
7 Use a Calculator to Evaluate Exponents (p. 24)
8 Use Scientific Notation (p. 24)
The Real Number Line
2 units
Scale
1 unit
O
3
2
1 1–2 0 –12 1 2 2
3
Figure 8 Real number line
DEFINITION
Real numbers can be represented by points on a line called the real number line.
There is a one-to-one correspondence between real numbers and points on a line.
That is, every real number corresponds to a point on the line, and each point on the
line has a unique real number associated with it.
Pick a point on a line somewhere in the center, and label it O. This point,
called the origin, corresponds to the real number 0. See Figure 8. The point 1 unit
to the right of O corresponds to the number 1. The distance between 0 and 1
determines the scale of the number line. For example, the point associated with the
number 2 is twice as far from O as 1. Notice that an arrowhead on the right end of
the line indicates the direction in which the numbers increase. Points to the left
of the origin correspond to the real numbers - 1, - 2, and so on. Figure 8 also shows
1
1
the points associated with the rational numbers - and and with the irrational
2
2
numbers 12 and p.
The real number associated with a point P is called the coordinate of P, and the
line whose points have been assigned coordinates is called the real number line.
Now Work
PROBLEM
13
18
CHAPTER R Review
O
3
2 3–2 1 1–2
Negative
real numbers
0
1
–
2
Zero
1
3–
2
2
3
Positive
real numbers
Figure 9
The real number line consists of three classes of real numbers, as shown in
Figure 9.
1. The negative real numbers are the coordinates of points to the left of the
origin O.
2. The real number zero is the coordinate of the origin O.
3. The positive real numbers are the coordinates of points to the right of the
origin O.
Multiplication Properties of Positive and Negative Numbers
1. The product of two positive numbers is a positive number.
2. The product of two negative numbers is a positive number.
3. The product of a positive number and a negative number is a negative
number.
1 Graph Inequalities
a
(a) a b
b
a
b
(b) a b
b
(c) a b
a
Figure 10
EX AMPLE 1
An important property of the real number line follows from the fact that, given two
numbers (points) a and b, either a is to the left of b, or a is at the same location
as b, or a is to the right of b. See Figure 10.
If a is to the left of b, then “a is less than b,” which is written a 6 b. If a is to
the right of b, then “a is greater than b,” which is written a 7 b. If a is at the same
location as b, then a = b. If a is either less than or equal to b, then a … b. Similarly,
a Ú b means that a is either greater than or equal to b. Collectively, the symbols
6 , 7 , … , and Ú are called inequality symbols.
Note that a 6 b and b 7 a mean the same thing. It does not matter whether
we write 2 6 3 or 3 7 2.
Furthermore, if a 6 b or if b 7 a, then the difference b - a is positive. Do you
see why?
Using Inequality Symbols
(a) 3 6 7
(d) - 8 6 - 4
(b) - 8 7 - 16
(e) 4 7 - 1
(c) - 6 6 0
(f) 8 7 0
r
In Example 1(a), we conclude that 3 6 7 either because 3 is to the left of 7 on
the real number line or because the difference, 7 - 3 = 4, is a positive real number.
Similarly, we conclude in Example 1(b) that - 8 7 - 16 either because
- 8 lies to the right of - 16 on the real number line or because the difference,
- 8 - 1 - 162 = - 8 + 16 = 8, is a positive real number.
Look again at Example 1. Note that the inequality symbol always points in the
direction of the smaller number.
An inequality is a statement in which two expressions are related by an inequality
symbol. The expressions are referred to as the sides of the inequality. Inequalities
of the form a 6 b or b 7 a are called strict inequalities, whereas inequalities of the
form a … b or b Ú a are called nonstrict inequalities.
Based on the discussion so far, we conclude that
a 7 0 is equivalent to a is positive
a 6 0 is equivalent to a is negative
SECTION R.2 Algebra Essentials
19
We sometimes read a 7 0 by saying that “a is positive.” If a Ú 0, then either a 7 0
or a = 0, and we may read this as “a is nonnegative.”
Now Work
PROBLEMS
17
AND
27
Graphing Inequalities
EXAMPL E 2
(a) On the real number line, graph all numbers x for which x 7 4.
(b) On the real number line, graph all numbers x for which x … 5.
Solution
–2 –1
0
1
2
3
4
5
6
7
3
4
5
6
7
(a) See Figure 11. Notice that we use a left parenthesis to indicate that the
number 4 is not part of the graph.
(b) See Figure 12. Notice that we use a right bracket to indicate that the number 5 is
part of the graph.
r
Figure 11 x 7 4
Now Work
2 1
0
1
2
PROBLEM
33
Figure 12 x … 5
2 Find Distance on the Real Number Line
4 units
3 units
5 4 3 2 1
0
1
2
3
4
The absolute value of a number a is the distance from 0 to a on the number line.
For example, - 4 is 4 units from 0, and 3 is 3 units from 0. See Figure 13. That is, the
absolute value of - 4 is 4, and the absolute value of 3 is 3.
A more formal definition of absolute value is given next.
Figure 13
DEFINITION
The absolute value of a real number a, denoted by the symbol 0 a 0 , is defined
by the rules
0 a 0 = a if a Ú 0
and
0 a 0 = - a if a 6 0
For example, because - 4 6 0, the second rule must be used to get
0 - 4 0 = - 1 - 42 = 4.
EXAMPL E 3
Computing Absolute Value
(a) 0 8 0 = 8
(b) 0 0 0 = 0
(c) 0 - 15 0 = - 1 - 152 = 15
r
Look again at Figure 13. The distance from - 4 to 3 is 7 units. This distance
is the difference 3 - 1 - 42, obtained by subtracting the smaller coordinate from
the larger. However, since 0 3 - 1 - 42 0 = 0 7 0 = 7 and 0 - 4 - 3 0 = 0 - 7 0 = 7, we
can use absolute value to calculate the distance between two points without being
concerned about which is smaller.
DEFINITION
If P and Q are two points on a real number line with coordinates a and b,
respectively, the distance between P and Q, denoted by d 1P, Q2, is
d 1P, Q2 = 0 b - a 0
Since 0 b - a 0 = 0 a - b 0 , it follows that d 1P, Q2 = d 1Q, P2.
20
CHAPTER R Review
EX AMPLE 4
Finding Distance on a Number Line
Let P, Q, and R be points on a real number line with coordinates - 5, 7, and - 3,
respectively. Find the distance
(a) between P and Q
Solution
(b) between Q and R
See Figure 14.
P
R
Q
5 4 3 2 1
0
1
2
3
4
5
6
7
d (P, Q) ⏐7 (5)⏐ 12
d(Q, R) ⏐ 3 7 ⏐ 10
Figure 14
(a) d 1P, Q2 = 0 7 - 1 - 52 0 = 0 12 0 = 12
(b) d 1Q, R2 = 0 - 3 - 7 0 = 0 - 10 0 = 10
Now Work
PROBLEM
r
39
3 Evaluate Algebraic Expressions
Remember, in algebra we use letters such as x, y, a, b, and c to represent numbers. If
the letter used is to represent any number from a given set of numbers, it is called
a variable. A constant is either a fixed number, such as 5 or 13, or a letter that
represents a fixed (possibly unspecified) number.
Constants and variables are combined using the operations of addition,
subtraction, multiplication, and division to form algebraic expressions. Examples of
algebraic expressions include
3
7x - 2y
1 - t
To evaluate an algebraic expression, substitute a numerical value for each
variable.
x + 3
EX AMPLE 5
Evaluating an Algebraic Expression
Evaluate each expression if x = 3 and y = - 1.
(a) x + 3y
Solution
(b) 5xy
(c)
(d) 0 - 4x + y 0
3y
2 - 2x
(a) Substitute 3 for x and - 1 for y in the expression x + 3y.
x + 3y = 3 + 31 - 12 = 3 + 1 - 32 = 0
c
x = 3, y = - 1
(b) If x = 3 and y = - 1, then
5xy = 5132 1 - 12 = - 15
(c) If x = 3 and y = - 1, then
31 - 12
3y
-3
-3
3
=
=
=
=
2 - 2x
2 - 2132
2 - 6
-4
4
(d) If x = 3 and y = - 1, then
0 - 4x + y 0 = 0 - 4132 + 1 - 12 0 = 0 - 12 + 1 - 12 0 = 0 - 13 0 = 13
Now Work
PROBLEMS
41
AND
49
r
SECTION R.2 Algebra Essentials
21
4 Determine the Domain of a Variable
In working with expressions or formulas involving variables, the variables may be
allowed to take on values from only a certain set of numbers. For example, in the
formula for the area A of a circle of radius r, A = pr 2, the variable r is necessarily
1
restricted to the positive real numbers. In the expression , the variable x cannot
x
take on the value 0, since division by 0 is not defined.
DEFINITION
EXAMPL E 6
The set of values that a variable may assume is called the domain of the
variable.
Finding the Domain of a Variable
The domain of the variable x in the expression
5
x - 2
is 5 x 0 x ≠ 26 since, if x = 2, the denominator becomes 0, which is not defined.
EXAMPL E 7
r
Circumference of a Circle
In the formula for the circumference C of a circle of radius r,
C = 2pr
the domain of the variable r, representing the radius of the circle, is the set of
positive real numbers, 5 r 0 r 7 06 . The domain of the variable C, representing the
circumference of the circle, is also the set of positive real numbers, 5 C 0 C 7 06 .
r
In describing the domain of a variable, we may use either set notation or words,
whichever is more convenient.
Now Work
PROBLEM
59
5 Use the Laws of Exponents
Integer exponents provide a shorthand notation for representing repeated
multiplications of a real number. For example,
34 = 3 # 3 # 3 # 3 = 81
Additionally, many formulas have exponents. For example,
r The formula for the horsepower rating H of an engine is
H =
D2 N
2.5
where D is the diameter of a cylinder and N is the number of cylinders.
r A formula for the resistance R of blood flowing in a blood vessel is
R = C
L
r4
where L is the length of the blood vessel, r is the radius, and C is a positive
constant.
22
CHAPTER R Review
DEFINITION
If a is a real number and n is a positive integer, then the symbol an represents
the product of n factors of a. That is,
an = a # a # c # a
u
(1)
n factors
In the definition it is understood that a1 = a. Furthermore, a2 = a # a, a3 = a # a # a,
and so on. In the expression an, a is called the base and n is called the exponent, or
power. We read an as “a raised to the power n” or as “a to the nth power.” We usually
- 24 = - 1 # 24 = - 16
read a2 as “a squared” and a3 as “a cubed.”
whereas
In working with exponents, the operation of raising to a power is performed
1- 22 4 = 1- 221- 22 1- 22 1- 22 = 16 before any other operation. As examples,
WARNING Be careful with minus signs
and exponents.
■
4 # 32 = 4 # 9 = 36
22 + 32 = 4 + 9 = 13
- 24 = - 16
5 # 32 + 2 # 4 = 5 # 9 + 2 # 4 = 45 + 8 = 53
Parentheses are used to indicate operations to be performed first. For example,
1 - 22 4 = 1 - 22 1 - 22 1 - 22 1 - 22 = 16
DEFINITION
12 + 32 2 = 52 = 25
If a ≠ 0, then
a0 = 1
DEFINITION
If a ≠ 0 and if n is a positive integer, then
a-n =
1
an
Whenever you encounter a negative exponent, think “reciprocal.”
EX AMPLE 8
Evaluating Expressions Containing Negative Exponents
(a) 2-3 =
1
1
=
8
23
Now Work
(b) x -4 =
PROBLEMS
1 -2
(c) a b =
5
1
x4
77
AND
1
1 2
a b
5
=
1
= 25
1
25
r
97
The following properties, called the Laws of Exponents, can be proved using the
preceding definitions. In the list, a and b are real numbers, and m and n are integers.
THEOREM
Laws of Exponents
am an = am + n
1am 2 = amn
1
am
m-n
= n-m
n = a
a
a
n
if a ≠ 0
1ab2 n = an bn
a n
an
a b = n
b
b
if b ≠ 0
SECTION R.2 Algebra Essentials
EXAMPL E 9
Using the Laws of Exponents
(a) x -3 # x5 = x -3 + 5 = x2 x ≠ 0
(b) 1x -3 2 = x -3 2 = x -6 =
(c) 12x2 3 = 23 # x3 = 8x3
2 4
24
16
(d) a b = 4 =
3
81
3
(e)
2
#
1
x6
x ≠ 0
x -2
= x -2 - 1-52 = x3 x ≠ 0
x -5
Now Work
EX AM PL E 1 0
23
PROBLEM
r
79
Using the Laws of Exponents
Write each expression so that all exponents are positive.
(a)
Solution
(a)
x5 y -2
x3 y
x5 y -2
x3 y
(b) ¢
x ≠ 0, y ≠ 0
=
(b) ¢
x -3 -2
≤
x ≠ 0, y ≠ 0
3y -1
x2
x5 # y -2
5 - 3 # -2 - 1
2 -3
2# 1
y
=
x
y
=
x
=
=
x
x3 y
y3
y3
1x -3 2 -2
x -3 -2
x6
x6
9x6
≤
=
=
=
=
1 2
y2
3y -1
13y -1 2 -2
3-2 1y -1 2 -2
y
9
Now Work
PROBLEM
r
89
6 Evaluate Square Roots
In Words
136 means “give me the
nonnegative number whose
square is 36.”
DEFINITION
A real number is squared when it is raised to the power 2. The inverse of squaring
is finding a square root. For example, since 62 = 36 and 1 - 62 2 = 36, the numbers
6 and - 6 are square roots of 36.
The symbol 1 , called a radical sign, is used to denote the principal, or
nonnegative, square root. For example, 136 = 6.
If a is a nonnegative real number, the nonnegative number b such that b2 = a
is the principal square root of a, and is denoted by b = 1a.
The following comments are noteworthy:
1. Negative numbers do not have square roots (in the real number system),
because the square of any real number is nonnegative. For example, 1- 4 is not
a real number, because there is no real number whose square is - 4.
2. The principal square root of 0 is 0, since 02 = 0. That is, 10 = 0.
3. The principal square root of a positive number is positive.
4. If c Ú 0, then 1 1c2 2 = c. For example, 1 122 2 = 2 and 1 132 2 = 3.
EX AM PL E 11
Evaluating Square Roots
(a) 264 = 8
(b)
1
1
=
4
A 16
(c) 1 21.42 2 = 1.4
r
Examples 11(a) and (b) are examples of square roots of perfect squares, since
1
1 2
64 = 82 and
= a b .
16
4
24
CHAPTER R Review
Consider the expression 2a2. Since a2 Ú 0, the principal square root of a2
is defined whether a 7 0 or a 6 0. However, since the principal square root is
nonnegative, we need an absolute value to ensure the nonnegative result. That is,
2a2 = 0 a 0
EX A MPL E 1 2
a any real number
Using Equation (2)
(a) 2 12.32 2 = 0 2.3 0 = 2.3
(2)
(b) 2 1 - 2.32 2 = 0 - 2.3 0 = 2.3
(c) 2x2 = 0 x 0
r
Now Work
PROBLEM
85
7 Use a Calculator to Evaluate Exponents
Your calculator has either a caret key, ^ , or an xy key, that is used for computations
involving exponents.
EX A MPL E 1 3
Exponents on a Graphing Calculator
Evaluate: 12.32 5
Solution Figure 15 shows the result using a TI-84 Plus C graphing calculator.
Now Work
PROBLEM
r
115
8 Use Scientific Notation
Measurements of physical quantities can range from very small to very large. For
example, the mass of a proton is approximately 0.00000000000000000000000000167
kilogram and the mass of Earth is about 5,980,000,000,000,000,000,000,000
kilograms. These numbers obviously are tedious to write down and difficult to read,
so we use exponents to rewrite them.
Figure 15
DEFINITION
When a number has been written as the product of a number x, where
1 … x 6 10, times a power of 10, it is said to be written in scientific notation.
In scientific notation,
Mass of a proton = 1.67 * 10-27 kilogram
Mass of Earth = 5.98 * 1024 kilograms
Converting a Decimal to Scientific Notation
To change a positive number into scientific notation:
1. Count the number N of places that the decimal point must be moved to
arrive at a number x, where 1 … x 6 10.
2. If the original number is greater than or equal to 1, the scientific notation is
x * 10N. If the original number is between 0 and 1, the scientific notation
is x * 10-N.
SECTION R.2 Algebra Essentials
EX AM PL E 1 4
25
Using Scientific Notation
Write each number in scientific notation.
(a) 9582
Solution
(b) 1.245
(c) 0.285
(d) 0.000561
(a) The decimal point in 9582 follows the 2. Count left from the decimal point
9
5
c
8
c
2
#
c
3
2 1
stopping after three moves, because 9.582 is a number between 1 and 10. Since
9582 is greater than 1, we write
9582 = 9.582 * 103
(b) The decimal point in 1.245 is between the 1 and 2. Since the number is already
between 1 and 10, the scientific notation for it is 1.245 * 100 = 1.245.
(c) The decimal point in 0.285 is between the 0 and the 2. We count
0
2
#
8
5
c
1
stopping after one move, because 2.85 is a number between 1 and 10. Since
0.285 is between 0 and 1, we write
0.285 = 2.85 * 10-1
(d) The decimal point in 0.000561 is moved as follows:
0
0
#
0
c
1
0
c
2
5
c
3
6
1
c
4
As a result,
r
0.000561 = 5.61 * 10-4
Now Work
EX AM PL E 15
121
PROBLEM
Changing from Scientific Notation to Decimals
Write each number as a decimal.
(a) 2.1 * 104
Solution
(b) 3.26 * 10-5
(a) 2.1 * 104 = 2
1
#
0
c
(b) 3.26 * 10-5 = 0
(c) 1 * 10
= 0
0
c
c
1
2
3
4
0
0
0
c
5
1
c
4
* 104 = 21,000
0
c
0
c
-2
0
c
c
3
#
Now Work
#
2
6 * 10-5 = 0.0000326
c
1
0 * 10
-2
= 0.01
r
1
PROBLEM
3
2
c
2
(c) 1 * 10-2
129
26
CHAPTER R Review
EX A MPL E 1 6
Using Scientific Notation
(a) The diameter of the smallest living cell is only about 0.00001 centimeter (cm).*
Express this number in scientific notation.
(b) The surface area of Earth is about 1.97 * 108 square miles.† Express the surface
area as a whole number.
Solution
(a) 0.00001 cm = 1 * 10-5 cm because the decimal point is moved five places and
the number is less than 1.
(b) 1.97 * 108 square miles = 197,000,000 square miles.
r
Now Work
PROBLEM
155
COMMENT On a calculator, a number such as 3.615 * 1012 is usually displayed as 3.615E12.
■
* Powers of Ten, Philip and Phylis Morrison.
†
2011 Information Please Almanac.
Historical Feature
T
he word algebra is derived from the Arabic word al-jabr. This
word is a part of the title of a ninth-century work, “Hisâb al-jabr
w’al-muqâbalah,” written by Mohammed ibn Músâ al-Khwârizmî.
The word al-jabr means “a restoration,” a reference to the fact that if a
number is added to one side of an equation, then it must also be added
to the other side in order to “restore” the equality. The title of the work,
freely translated, is“The Science of Reduction and Cancellation.”Of course,
today, algebra has come to mean a great deal more.
R.2 Assess Your Understanding
Concepts and Vocabulary
1. A(n)
is a letter used in algebra to represent any
number from a given set of numbers.
2. On the real number line, the real number zero is the
coordinate of the
.
3. An inequality of the form a 7 b is called a(n)
inequality.
4. In the expression 24, the number 2 is called the
is called the
.
5. In scientific notation, 1234.5678 =
and 4
.
6. If a is a nonnegative real number, then which inequality
statement best describes a?
(a) a 6 0
(b) a 7 0
(c) a … 0
(d) a Ú 0
7. Let a and b be non-zero real numbers and m and n be
integers. Which of the following is not a law of exponents?
an
a n
(a) a b = n
b
b
am
(c) n = am - n
a
(b) 1am 2 n = am + n
(d) 1ab2 n = anbn
8. True or False The product of two negative real numbers is
always greater than zero.
9. True or False The distance between two distinct points on
the real number line is always greater than zero.
10. True or False The absolute value of a real number is always
greater than zero.
11. True or False When a number is expressed in scientific
notation, it is expressed as the product of a number x,
0 … x 6 1, and a power of 10.
12. True or False To multiply two expressions having the same
base, retain the base and multiply the exponents.
Skill Building
5
3
13. On the real number line, label the points with coordinates 0, 1, - 1, , - 2.5, , and 0.25.
2
4
3 1
2
14. Repeat Problem 13 for the coordinates 0, - 2, 2, - 1.5, , , and .
2 3
3
In Problems 15–24, replace the question mark by 6, 7, or =, whichever is correct.
15.
1
?0
2
20. 22 ? 1.41
16. 5 ? 6
21.
1
? 0.5
2
17. - 1 ? - 2
22.
1
? 0.33
3
18. - 3 ? 23.
2
? 0.67
3
5
2
19. p ? 3.14
24.
1
? 0.25
4
SECTION R.2 Algebra Essentials
27
In Problems 25–30, write each statement as an inequality.
25. x is positive
26. z is negative
27. x is less than 2
28. y is greater than - 5
29. x is less than or equal to 1
30. x is greater than or equal to 2
In Problems 31–34, graph the numbers x on the real number line.
32. x 6 4
31. x Ú - 2
34. x … 7
33. x 7 - 1
In Problems 35–40, use the given real number line to compute each distance.
A
35. d1C, D2
B
C
D
4 3 2 1
0
1
36. d1C, A2
E
2
37. d1D, E2
3
4
5
6
38. d1C, E2
39. d1A, E2
40. d1D, B2
In Problems 41–48, evaluate each expression if x = - 2 and y = 3.
41. x + 2y
45.
42. 3x + y
2x
x - y
46.
x + y
x - y
47.
3x + 2y
2 + y
In Problems 49–58, find the value of each expression if x = 3 and y = - 2.
49. 0 x + y 0
54.
0y0
y
44. - 2x + xy
43. 5xy + 2
48.
2x - 3
y
0x0
50. 0 x - y 0
51. 0 x 0 + 0 y 0
52. 0 x 0 - 0 y 0
53.
55. 0 4x - 5y 0
56. 0 3x + 2y 0
57. 0 0 4x 0 - 0 5y 0 0
58. 3 0 x 0 + 2 0 y 0
x
In Problems 59–66, determine which of the values (a) through (d), if any, must be excluded from the domain of the variable in each
expression.
(a) x = 3
(b) x = 1
(c) x = 0
(d) x = - 1
59.
x2 - 1
x
60.
x2 + 1
x
61.
x
x2 - 9
62.
x
x2 + 9
63.
x2
x + 1
64.
x3
x - 1
65.
x2 + 5x - 10
x3 - x
66.
- 9x2 - x + 1
x3 + x
70.
x - 2
x - 6
2
2
In Problems 67–70, determine the domain of the variable x in each expression.
x
4
-6
69.
67.
68.
x
+
4
x - 5
x + 4
5
In Problems 71–74, use the formula C = 1F - 322 for converting degrees Fahrenheit into degrees Celsius to find the Celsius measure
9
of each Fahrenheit temperature.
71. F = 32°
72. F = 212°
73. F = 77°
74. F = - 4°
In Problems 75–86, simplify each expression.
75. 1 - 42 2
76. - 42
77. 4-2
78. - 4-2
79. 3-6 # 34
80. 4-2 # 43
81. 13-2 2 -1
82. 12-1 2 -3
83. 225
84. 236
85. 21 - 42 2
86. 21 - 32 2
In Problems 87–96, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative,
we assume that the base is not 0.
87. 18x3 2
92.
x -2 y
xy2
2
88. 1 - 4x2 2 - 1
93.
1 - 22 3 x4 1yz2 2
32 xy3 z
89. 1x2 y -1 2
94.
2
4x -2 1yz2 -1
23 x4 y
90. 1x -1 y2
95. ¢
3x -1
4y
3
≤
-1
-2
91.
x2 y 3
xy4
96. ¢
5x -2 -3
≤
6y -2
28
CHAPTER R Review
In Problems 97–108, find the value of each expression if x = 2 and y = - 1.
99. x2 + y2
100. x2 y2
97. 2xy -1
98. - 3x -1 y
101. 1xy2 2
102. 1x + y2 2
103. 2x2
104. 1 1x2 2
105. 2x2 + y2
106. 2x2 + 2y2
107. xy
108. yx
109. Find the value of the expression 2x3 - 3x2 + 5x - 4 if x = 2. What is the value if x = 1?
110. Find the value of the expression 4x3 + 3x2 - x + 2 if x = 1. What is the value if x = 2?
111. What is the value of
16662 4
12222 4
112. What is the value of 10.12 3 1202 3 ?
?
In Problems 113–120, use a calculator to evaluate each expression. Round your answer to three decimal places.
113. 18.22 6
114. 13.72 5
115. 16.12 -3
116. 12.22 -5
117. 1 - 2.82 6
118. - 12.82 6
119. 1 - 8.112 -4
120. - 18.112 -4
In Problems 121–128, write each number in scientific notation.
121. 454.2
122. 32.14
123. 0.013
124. 0.00421
125. 32,155
126. 21,210
127. 0.000423
128. 0.0514
In Problems 129–136, write each number as a decimal.
129. 6.15 * 104
130. 9.7 * 103
131. 1.214 * 10-3
132. 9.88 * 10-4
133. 1.1 * 108
134. 4.112 * 102
135. 8.1 * 10-2
136. 6.453 * 10-1
Applications and Extensions
In Problems 137–146, express each statement as an equation involving the indicated variables.
137. Area of a Rectangle The area A of a rectangle is the
product of its length l and its width w.
l
A
141. Area of an Equilateral Triangle The area A of an
13
equilateral triangle is
times the square of the length x of
4
one side.
w
x
x
138. Perimeter of a Rectangle The perimeter P of a rectangle is
twice the sum of its length l and its width w.
139. Circumference of a Circle The circumference C of a circle is
the product of p and its diameter d.
C
x
142. Perimeter of an Equilateral Triangle The perimeter P of an
equilateral triangle is 3 times the length x of one side.
4
143. Volume of a Sphere The volume V of a sphere is times p
3
times the cube of the radius r.
d
r
140. Area of a Triangle The area A of a triangle is one-half the
product of its base b and its height h.
h
b
144. Surface Area of a Sphere The surface area S of a sphere is
4 times p times the square of the radius r.
SECTION R.2 Algebra Essentials
145. Volume of a Cube The volume V of a cube is the cube of
the length x of a side.
of its products is a ball bearing with a stated radius of
3 centimeters (cm). Only ball bearings with a radius within
0.01 cm of this stated radius are acceptable. If x is the radius
of a ball bearing, a formula describing this situation is
0 x - 3 0 … 0.01
x
x
x
146. Surface Area of a Cube The surface area S of a cube is 6
times the square of the length x of a side.
147. Manufacturing Cost The weekly production cost C of
manufacturing x watches is given by the formula
C = 4000 + 2x, where the variable C is in dollars.
(a) What is the cost of producing 1000 watches?
(b) What is the cost of producing 2000 watches?
148. Balancing a Checkbook At the beginning of the month,
Mike had a balance of $210 in his checking account. During
the next month, he deposited $80, wrote a check for $120,
made another deposit of $25, and wrote two checks: one for
$60 and the other for $32. He was also assessed a monthly
service charge of $5. What was his balance at the end of the
month?
29
(a) Is a ball bearing of radius x = 2.999 acceptable?
(b) Is a ball bearing of radius x = 2.89 acceptable?
154. Body Temperature Normal human body temperature is
98.6°F. A temperature x that differs from normal by at least
1.5°F is considered unhealthy. A formula that describes this is
0 x - 98.6 0 Ú 1.5
(a) Show that a temperature of 97°F is unhealthy.
(b) Show that a temperature of 100°F is not unhealthy.
155. Distance from Earth to Its Moon The distance from Earth
to the Moon is about 4 * 108 meters.* Express this distance
as a whole number.
156. Height of Mt. Everest The height of Mt. Everest is 8848
meters.* Express this height in scientific notation.
157. Wavelength of Visible Light The wavelength of visible
light is about 5 * 10-7 meter.* Express this wavelength as a
decimal.
In Problems 149 and 150, write an inequality using an absolute
value to describe each statement.
158. Diameter of an Atom The diameter of an atom is about
1 * 10-10 meter.* Express this diameter as a decimal.
149. x is at least 6 units from 4.
159. Diameter of Copper Wire The smallest commercial copper
wire is about 0.0005 inch in diameter.† Express this diameter
using scientific notation.
150. x is more than 5 units from 2.
151. U.S. Voltage In the United States, normal household
voltage is 110 volts. It is acceptable for the actual voltage x
to differ from normal by at most 5 volts. A formula that
describes this is
0 x - 110 0 … 5
(a) Show that a voltage of 108 volts is acceptable.
(b) Show that a voltage of 104 volts is not acceptable.
152. Foreign Voltage In other countries, normal household
voltage is 220 volts. It is acceptable for the actual voltage x
to differ from normal by at most 8 volts. A formula that
describes this is
0 x - 220 0 … 8
(a) Show that a voltage of 214 volts is acceptable.
(b) Show that a voltage of 209 volts is not acceptable.
153. Making Precision Ball Bearings The FireBall Company
manufactures ball bearings for precision equipment. One
160. Smallest Motor The smallest motor ever made is less than
0.05 centimeter wide.† Express this width using scientific
notation.
161. Astronomy One light-year is defined by astronomers to
be the distance that a beam of light will travel in 1 year
(365 days). If the speed of light is 186,000 miles per second,
how many miles are in a light-year? Express your answer in
scientific notation.
162. Astronomy How long does it take a beam of light to reach
Earth from the Sun when the Sun is 93,000,000 miles from
Earth? Express your answer in seconds, using scientific
notation.
163. Does
1
equal 0.333? If not, which is larger? By how much?
3
164. Does
2
equal 0.666? If not, which is larger? By how much?
3
Explaining Concepts: Discussion and Writing
165. Is there a positive real number “closest” to 0?
166. Number game I’m thinking of a number! It lies between 1
and 10; its square is rational and lies between 1 and 10. The
number is larger than p. Correct to two decimal places
(that is, truncated to two decimal places), name the number.
Now think of your own number, describe it, and challenge a
fellow student to name it.
*
†
Powers of Ten, Philip and Phylis Morrison.
2011 Information Please Almanac.
167. Write a brief paragraph that illustrates the similarities and
differences between “less than” 1 6 2 and “less than or equal
to” 1 … 2.
168. Give a reason why the statement 5 6 8 is true.
30
CHAPTER R Review
R.3 Geometry Essentials
OBJECTIVES 1 Use the Pythagorean Theorem and Its Converse (p. 30)
2 Know Geometry Formulas (p. 31)
3 Understand Congruent Triangles and Similar Triangles (p. 32)
1 Use the Pythagorean Theorem and Its Converse
Hypotenuse
c
b
Leg
90°
a
Leg
The Pythagorean Theorem is a statement about right triangles. A right triangle is one
that contains a right angle—that is, an angle of 90°. The side of the triangle opposite
the 90° angle is called the hypotenuse; the remaining two sides are called legs. In
Figure 16 we have used c to represent the length of the hypotenuse and a and b to
represent the lengths of the legs. Notice the use of the symbol
to show the 90° angle.
We now state the Pythagorean Theorem.
Figure 16 A right triangle
PYTHAGOREAN
THEOREM
In a right triangle, the square of the length of the hypotenuse is equal to the
sum of the squares of the lengths of the legs. That is, in the right triangle shown
in Figure 16,
c 2 = a 2 + b2
(1)
A proof of the Pythagorean Theorem is given at the end of this section.
EX AMPLE 1
Finding the Hypotenuse of a Right Triangle
In a right triangle, one leg has length 4 and the other has length 3. What is the length
of the hypotenuse?
Solution
Since the triangle is a right triangle, we use the Pythagorean Theorem with a = 4
and b = 3 to find the length c of the hypotenuse. From equation (1),
c 2 = a 2 + b2
c 2 = 42 + 32 = 16 + 9 = 25
r
c = 225 = 5
Now Work
PROBLEM
15
The converse of the Pythagorean Theorem is also true.
CONVERSE OF THE
PYTHAGOREAN
THEOREM
In a triangle, if the square of the length of one side equals the sum of the
squares of the lengths of the other two sides, the triangle is a right triangle. The
90° angle is opposite the longest side.
A proof is given at the end of this section.
EX A MPLE 2
Verifying That a Triangle Is a Right Triangle
Show that a triangle whose sides are of lengths 5, 12, and 13 is a right triangle.
Identify the hypotenuse.
Solution
Square the lengths of the sides.
52 = 25
122 = 144
132 = 169
SECTION R.3 Geometry Essentials
31
Notice that the sum of the first two squares (25 and 144) equals the third square
(169). That is, because 52 + 122 = 132, the triangle is a right triangle. The longest
side, 13, is the hypotenuse. See Figure 17.
13
r
5
90°
Now Work
PROBLEM
12
23
Figure 17
Applying the Pythagorean Theorem
EXAMPL E 3
The tallest building in the world is Burj Khalifa in Dubai, United Arab Emirates, at
2717 feet and 163 floors. The observation deck is 1483 feet above ground level. How
far can a person standing on the observation deck see (with the aid of a telescope)?
Use 3960 miles for the radius of Earth.
Source: Council on Tall Buildings and Urban Habitat
Solution From the center of Earth, draw two radii: one through Burj Khalifa
and the other to the farthest point a person can see from the observation deck. See
Figure 18. Apply the Pythagorean Theorem to the right triangle.
1483
mile. Then
Since 1 mile = 5280 feet, 1483 feet =
5280
d 2 + 139602 2 = a3960 +
d 2 = a3960 +
1483 2
b
5280
1483 2
b - 139602 2 ≈ 2224.58
5280
d ≈ 47.17
A person can see more than 47 miles from the observation deck.
d
1483 ft
3960 mi
r
Figure 18
Now Work
PROBLEM
55
2 Know Geometry Formulas
Certain formulas from geometry are useful in solving algebra problems.
For a rectangle of length l and width w,
w
Area = lw
Perimeter = 2l + 2w
l
For a triangle with base b and altitude h,
h
b
Area =
1
bh
2
32
CHAPTER R Review
For a circle of radius r (diameter d = 2r),
d
r
Area = pr 2
Circumference = 2pr = pd
For a closed rectangular box of length l, width w, and height h,
h
w
l
Volume = lwh
Surface area = 2lh + 2wh + 2lw
For a sphere of radius r,
r
Volume =
4 3
pr
3
Surface area = 4pr 2
For a closed right circular cylinder of height h and radius r,
r
h
Volume = pr 2 h
Now Work
EX AMPLE 4
PROBLEM
Surface area = 2pr 2 + 2prh
31
Using Geometry Formulas
A Christmas tree ornament is in the shape of a semicircle on top of a triangle. How
many square centimeters (cm2) of copper is required to make the ornament if the
height of the triangle is 6 cm and the base is 4 cm?
Solution
4
l6
See Figure 19. The amount of copper required equals the shaded area. This area is
the sum of the areas of the triangle and the semicircle. The triangle has height h = 6
and base b = 4. The semicircle has diameter d = 4, so its radius is r = 2.
Area = Area of triangle + Area of semicircle
=
Figure 19
1
1
1
1
bh + pr 2 = 142 162 + p # 22
2
2
2
2
b = 4; h = 6; r = 2
= 12 + 2p ≈ 18.28 cm2
About 18.28 cm2 of copper is required.
Now Work
PROBLEM
r
49
3 Understand Congruent Triangles and Similar Triangles
In Words
Two triangles are congruent if
they have the same size and
shape.
DEFINITION
Throughout the text we will make reference to triangles. We begin with a discussion
of congruent triangles. According to dictionary.com, the word congruent means
“coinciding exactly when superimposed.” For example, two angles are congruent if
they have the same measure, and two line segments are congruent if they have the
same length.
Two triangles are congruent if each pair of corresponding angles have the same
measure and each pair of corresponding sides are the same length.
In Figure 20, corresponding angles are equal and the corresponding sides are
equal in length: a = d, b = e, and c = f . As a result, these triangles are congruent.
SECTION R.3 Geometry Essentials
100
a
30
100
d
b
e
50
30
50
c
Figure 20
33
f
Congruent triangles
Actually, it is not necessary to verify that all three angles and all three sides are
the same measure to determine whether two triangles are congruent.
Determining Congruent Triangles
1. Angle–Side–Angle Case Two triangles are congruent if two of the angles
are equal and the lengths of the corresponding sides between the two angles
are equal.
For example, in Figure 21(a), the two triangles are congruent because
two angles and the included side are equal.
2. Side–Side–Side Case Two triangles are congruent if the lengths of the
corresponding sides of the triangles are equal.
For example, in Figure 21(b), the two triangles are congruent because
the three corresponding sides are all equal.
3. Side–Angle–Side Case Two triangles are congruent if the lengths of two
corresponding sides are equal and the angles between the two sides are the
same.
For example, in Figure 21(c), the two triangles are congruent because
two sides and the included angle are equal.
15
20
80
80
10
15
20
40
(a)
40
8
40
7
10
40
8
8
7
8
(b)
(c)
Figure 21
We contrast congruent triangles with similar triangles.
DEFINITION
In Words
Two triangles are similar if they
have the same shape, but
(possibly) different sizes.
Two triangles are similar if the corresponding angles are equal and the lengths
of the corresponding sides are proportional.
For example, the triangles in Figure 22 (on the next page) are similar because the
corresponding angles are equal. In addition, the lengths of the corresponding sides
are proportional because each side in the triangle on the right is twice as long as each
corresponding side in the triangle on the left. That is, the ratio of the corresponding
f
d
e
sides is a constant: =
= = 2.
a
c
b
34
CHAPTER R Review
80
d 5 2a
a
80
30
e 5 2b
b
30
70
70
f 5 2c
c
Figure 22 Similar triangles
It is not necessary to verify that all three angles are equal and all three sides are
proportional to determine whether two triangles are congruent.
Determining Similar Triangles
1. Angle–Angle Case Two triangles are similar if two of the corresponding
angles are equal.
For example, in Figure 23(a), the two triangles are similar because two
angles are equal.
2. Side–Side–Side Case Two triangles are similar if the lengths of all three
sides of each triangle are proportional.
For example, in Figure 23(b), the two triangles are similar because
10
5
6
1
=
=
=
30
15
18
3
3. Side–Angle–Side Case Two triangles are similar if two corresponding
sides are proportional and the angles between the two sides are equal.
For example, in Figure 23(c), the two triangles are similar because
4
12
2
=
= and the angles between the sides are equal.
6
18
3
15
80
80
30
5
10
18
120
120
6
35
35
18
12
6
4
(a)
(b)
(c)
Figure 23
EX AMPLE 5
Using Similar Triangles
Given that the triangles in Figure 24 are similar, find the missing length x and the
angles A, B, and C.
60
6
3
30
x
B
C
5
90
Figure 24
A
SECTION R.3 Geometry Essentials
Solution
35
Because the triangles are similar, corresponding angles are equal. So A = 90°,
B = 60°, and C = 30°. Also, the corresponding sides are proportional. That is,
3
6
= . We solve this equation for x.
x
5
3
6
=
x
5
3
6
5x # = 5x # Multiply both sides by 5x.
x
5
Simplify.
3x = 30
Divide both sides by 3.
x = 10
r
The missing length is 10 units.
Now Work
PROBLEM
43
Proof of the Pythagorean Theorem Begin with a square, each side of length
a + b. In this square, form four right triangles, each having legs equal in length to a
and b. See Figure 25. All these triangles are congruent (two sides and their included
angle are equal). As a result, the hypotenuse of each is the same, say c, and the pink
shading in Figure 25 indicates a square with an area equal to c 2.
b
Area
1
= –2 ab
a
1
Area = –2 ab
a
c
c
b
Area = c 2
c
b
c
1
Area = –2 ab
a
1
Area = –2 ab
a
b
Figure 25
The area of the original square with sides a + b equals the sum of the areas of the
1
four triangles (each of area ab) plus the area of the square with side c. That is,
2
1
1
1
1
1a + b2 2 = ab + ab + ab + ab + c 2
2
2
2
2
a2 + 2ab + b2 = 2ab + c 2
a 2 + b2 = c 2
The proof is complete.
x
b
a
■
Proof of the Converse of the Pythagorean Theorem Begin with two triangles:
one a right triangle with legs a and b and the other a triangle with sides a, b, and c
for which c 2 = a2 + b2. See Figure 26. By the Pythagorean Theorem, the length x
of the third side of the first triangle is
x 2 = a 2 + b2
(a)
But c 2 = a2 + b2. Then,
c
b
a
2
2
(b) c = a + b
Figure 26
2
x2 = c 2
x = c
The two triangles have the same sides and are therefore congruent. This means
corresponding angles are equal, so the angle opposite side c of the second triangle
equals 90°.
The proof is complete.
■
36
CHAPTER R Review
R.3 Assess Your Understanding
Concepts and Vocabulary
10. True or False The triangles shown are congruent.
1. A(n)
triangle is one that contains an angle of
90 degrees. The longest side is called the
.
10
2. For a triangle with base b and altitude h, a formula for the
area A is
30
.
30
3. The formula for the circumference C of a circle of radius r is
.
29
29
10
if corresponding angles are equal
4. Two triangles are
and the lengths of the corresponding sides are proportional.
11. True or False The triangles shown are similar.
5. Which of the following is not a case for determining
congruent triangles?
(a) Angle–Side–Angle
(b) Side–Angle–Side
(c) Angle–Angle–Angle
(d) Side-Side-Side
25
25
6. Choose the formula for the volume of a sphere of radius r.
4
4
(a) pr 2
(b) pr 3
(c) 4pr 3
(d) 4pr 2
3
3
100
7. True or False In a right triangle, the square of the length of
the longest side equals the sum of the squares of the lengths
of the other two sides.
100
12. True or False The triangles shown are similar.
8. True or False The triangle with sides of lengths 6, 8, and 10
is a right triangle.
9. True or False The surface area of a sphere of radius r is
4 2
pr .
3
4
3
120⬚
120⬚
2
3
Skill Building
In Problems 13–18, the lengths of the legs of a right triangle are given. Find the hypotenuse.
13. a = 5, b = 12
14. a = 6, b = 8
15. a = 10, b = 24
16. a = 4, b = 3
17. a = 7, b = 24
18. a = 14, b = 48
In Problems 19–26, the lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the
hypotenuse.
19. 3, 4, 5
20. 6, 8, 10
21. 4, 5, 6
22. 2, 2, 3
23. 7, 24, 25
24. 10, 24, 26
25. 6, 4, 3
26. 5, 4, 7
27. Find the area A of a rectangle with length 4 inches and width 2 inches.
28. Find the area A of a rectangle with length 9 centimeters and width 4 centimeters.
29. Find the area A of a triangle with height 4 inches and base 2 inches.
30. Find the area A of a triangle with height 9 centimeters and base 4 centimeters.
31. Find the area A and circumference C of a circle of radius 5 meters.
32. Find the area A and circumference C of a circle of radius 2 feet.
33. Find the volume V and surface area S of a closed rectangular box with length 8 feet, width 4 feet, and height 7 feet.
34. Find the volume V and surface area S of a closed rectangular box with length 9 inches, width 4 inches, and height 8 inches.
35. Find the volume V and surface area S of a sphere of radius 4 centimeters.
36. Find the volume V and surface area S of a sphere of radius 3 feet.
37. Find the volume V and surface area S of a closed right circular cylinder with radius 9 inches and height 8 inches.
38. Find the volume V and surface area S of a closed right circular cylinder with radius 8 inches and height 9 inches.
SECTION R.3 Geometry Essentials
37
In Problems 39–42, find the area of the shaded region.
39.
40.
2
41.
2
42.
2
2
2
2
2
2
In Problems 43–46, the triangles in each pair are similar. Find the missing length x and the missing angles A, B, and C.
43. 60
44.
2
30
16
90
45.
75
4
10
125
50
20
95
12
30
46.
60
25
50
45
75
A
A
x
B
x
A
6
5
30
A
B
8
8 B
C
x
B
C
x
C
C
Applications and Extensions
47. How many feet has a wheel with a diameter of 16 inches
traveled after four revolutions?
48. How many revolutions will a circular disk with a diameter of
4 feet have completed after it has rolled 20 feet?
51. Architecture A Norman window consists of a rectangle
surmounted by a semicircle. Find the area of the Norman
window shown in the illustration. How much wood frame is
needed to enclose the window?
49. In the figure shown, ABCD is a square, with each side of
length 6 feet. The width of the border (shaded portion)
between the outer square EFGH and ABCD is 2 feet. Find
the area of the border.
E
F
A
6'
B
6 ft
2
ft
D
4'
C
H
G
50. Refer to the figure. Square ABCD has an area of 100 square
feet; square BEFG has an area of 16 square feet. What is the
area of the triangle CGF ?
A
B
D
E
F
G
C
52. Construction A circular swimming pool that is 20 feet in
diameter is enclosed by a wooden deck that is 3 feet wide.
What is the area of the deck? How much fence is required to
enclose the deck?
3'
20'
38
CHAPTER R Review
53. How Tall Is the Great Pyramid? The ancient Greek
philosopher Thales of Miletus is reported on one occasion
to have visited Egypt and calculated the height of the Great
Pyramid of Cheops by means of shadow reckoning. Thales
knew that each side of the base of the pyramid was 252 paces
and that his own height was 2 paces. He measured the length
of the pyramid’s shadow to be 114 paces and determined the
length of his shadow to be 3 paces. See the illustration. Using
similar triangles, determine the height of the Great Pyramid
in terms of the number of paces.
54. The Bermuda Triangle Karen is doing research on the
Bermuda Triangle which she defines roughly by Hamilton,
Bermuda; San Juan, Puerto Rico; and Fort Lauderdale, Florida.
On her atlas Karen measures the straight-line distances from
Hamilton to Fort Lauderdale, Fort Lauderdale to San Juan,
and San Juan to Hamilton to be approximately 57 millimeters
(mm), 58 mm, and 53.5 mm respectively. If the actual
distance from Fort Lauderdale to San Juan is 1046 miles,
approximate the actual distances from San Juan to Hamilton
and from Hamilton to Fort Lauderdale.
Source: Diggins, Julie E, String Straightedge and Shadow:
The Story of Geometry, 2003, Whole Spirit Press,
http://wholespiritpress.com.
In Problems 55–57, use the facts that the radius of Earth is 3960 miles and 1 mile = 5280 feet.
55. How Far Can You See? The conning tower of the U.S.S.
Silversides, a World War II submarine now permanently
stationed in Muskegon, Michigan, is approximately 20 feet
above sea level. How far can you see from the conning tower?
56. How Far Can You See? A person who is 6 feet tall is standing
on the beach in Fort Lauderdale, Florida, and looks out onto
the Atlantic Ocean. Suddenly, a ship appears on the horizon.
How far is the ship from shore?
How far can a person see from the bridge, which is 150 feet
above sea level?
58. Suppose that m and n are positive integers with m 7 n. If
a = m2 - n2, b = 2mn, and c = m2 + n2, show that a, b,
and c are the lengths of the sides of a right triangle. (This
formula can be used to find the sides of a right triangle that
are integers, such as 3, 4, 5; 5, 12, 13; and so on. Such triplets
of integers are called Pythagorean triples.)
57. How Far Can You See? The deck of a destroyer is 100 feet
above sea level. How far can a person see from the deck?
Explaining Concepts: Discussion and Writing
59. You have 1000 feet of flexible pool siding and intend to
construct a swimming pool. Experiment with rectangular-shaped
pools with perimeters of 1000 feet. How do their areas vary?
What is the shape of the rectangle with the largest area?
Now compute the area enclosed by a circular pool with a
perimeter (circumference) of 1000 feet. What would be your
choice of shape for the pool? If rectangular, what is your
preference for dimensions? Justify your choice. If your only
consideration is to have a pool that encloses the most area,
what shape should you use?
60. The Gibb’s Hill Lighthouse, Southampton, Bermuda, in
operation since 1846, stands 117 feet high on a hill 245
feet high, so its beam of light is 362 feet above sea level. A
brochure states that the light itself can be seen on the horizon
about 26 miles distant. Verify the accuracy of this information.
The brochure further states that ships 40 miles away can
see the light and that planes flying at 10,000 feet can see
it 120 miles away. Verify the accuracy of these statements.
What assumption did the brochure make about the height of
the ship?
120 miles
40 miles
SECTION R.4 Polynomials
39
R.4 Polynomials
OBJECTIVES 1
2
3
4
5
6
7
Recognize Monomials (p. 39)
Recognize Polynomials (p. 40)
Add and Subtract Polynomials (p. 41)
Multiply Polynomials (p. 42)
Know Formulas for Special Products (p. 43)
Divide Polynomials Using Long Division (p. 44)
Work with Polynomials in Two Variables (p. 47)
We have described algebra as a generalization of arithmetic in which letters are
used to represent real numbers. From now on, we shall use the letters at the end
of the alphabet, such as x, y, and z, to represent variables and use the letters at the
beginning of the alphabet, such as a, b, and c, to represent constants. In the
expressions 3x + 5 and ax + b, it is understood that x is a variable and that a and
b are constants, even though the constants a and b are unspecified. As you will find
out, the context usually makes the intended meaning clear.
1 Recognize Monomials
DEFINITION
NOTE The nonnegative integers are the
integers 0, 1, 2, 3,….
■
A monomial in one variable is the product of a constant and a variable raised
to a nonnegative integer power. A monomial is of the form
axk
where a is a constant, x is a variable, and k Ú 0 is an integer. The constant a is
called the coefficient of the monomial. If a ≠ 0, then k is called the degree of
the monomial.
EXAMPL E 1
Examples of Monomials
Monomial
(a)
(b)
(c)
(d)
(e)
EXAMPL E 2
2
6x
- 22x3
3
- 5x
x4
Coefficient
Degree
6
- 22
3
-5
1
2
3
0
1
4
Since 3 = 3 # 1 = 3x 0, x ≠ 0
Since - 5x = - 5x 1
Since x 4 = 1 # x 4
r
Examples of Nonmonomial Expressions
1
1
(a) 3x1>2 is not a monomial, since the exponent of the variable x is , and is not a
2
2
nonnegative integer.
(b) 4x -3 is not a monomial, since the exponent of the variable x is - 3, and - 3 is not
a nonnegative integer.
r
Now Work
PROBLEM
9
40
CHAPTER R Review
2 Recognize Polynomials
Two monomials with the same variable raised to the same power are called like
terms. For example, 2x4 and - 5x4 are like terms. In contrast, the monomials 2x3 and
2x5 are not like terms.
We can add or subtract like terms using the Distributive Property. For example,
2x2 + 5x2 = 12 + 52x2 = 7x2 and 8x3 - 5x3 = 18 - 52x3 = 3x3
The sum or difference of two monomials having different degrees is called a
binomial. The sum or difference of three monomials with three different degrees is
called a trinomial. For example,
x2 - 2 is a binomial.
x3 - 3x + 5 is a trinomial.
2x2 + 5x2 + 2 = 7x2 + 2 is a binomial.
DEFINITION
A polynomial in one variable is an algebraic expression of the form
an xn + an - 1 xn - 1 + g + a1 x + a0
(1)
where an , an - 1 ,c, a1 , a0 are constants,* called the coefficients of the polynomial,
n Ú 0 is an integer, and x is a variable. If an ≠ 0, it is called the leading
coefficient, anxn is called the leading term, and n is the degree of the polynomial.
In Words
A polynomial is a sum of
monomials.
The monomials that make up a polynomial are called its terms. If all of the
coefficients are 0, the polynomial is called the zero polynomial, which has no degree.
Polynomials are usually written in standard form, beginning with the nonzero
term of highest degree and continuing with terms in descending order according to
degree. If a power of x is missing, it is because its coefficient is zero.
EX AMPLE 3
Examples of Polynomials
Polynomial
3
2
- 8x + 4x - 6x + 2
Coefficients
Degree
3x - 5 = 3x + 0 # x + 1 - 52
8 - 2x + x2 = 1 # x2 + 1 - 22x + 8
- 8, 4, - 6, 2
3
3, 0, - 5
2
1, - 2, 8
2
5x + 22 = 5x + 22
5, 22
1
3 = 3 # 1 = 3 # x0
3
0
0
0
No degree
2
2
1
r
Although we have been using x to represent the variable, letters such as y and
z are also commonly used.
3x4 - x2 + 2 is a polynomial (in x) of degree 4.
9y3 - 2y2 + y - 3 is a polynomial (in y) of degree 3.
z5 + p is a polynomial (in z) of degree 5.
Algebraic expressions such as
1
x
and
x2 + 1
x + 5
*The notation an is read as “a sub n.” The number n is called a subscript and should not be confused
with an exponent. We use subscripts to distinguish one constant from another when a large or
undetermined number of constants are required.
SECTION R.4 Polynomials
41
1
= x -1 has an exponent
x
that is not a nonnegative integer. Although the second expression is the quotient of
two polynomials, the polynomial in the denominator has degree greater than 0, so
the expression cannot be a polynomial.
are not polynomials. The first is not a polynomial because
Now Work
PROBLEM
19
3 Add and Subtract Polynomials
Polynomials are added and subtracted by combining like terms.
EXAMPL E 4
Adding Polynomials
Find the sum of the polynomials:
8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + x
Solution
We shall find the sum in two ways.
Horizontal Addition: The idea here is to group the like terms and then combine
them.
18x3 - 2x2 + 6x - 22 + 13x4 - 2x3 + x2 + x2
= 3x4 + 18x3 - 2x3 2 + 1 - 2x2 + x2 2 + 16x + x2 - 2
= 3x4 + 6x3 - x2 + 7x - 2
Vertical Addition: The idea here is to vertically line up the like terms in each
polynomial and then add the coefficients.
x4
x3
x2
3
x1
x0
2
8x - 2x + 6x - 2
+ 3x4 - 2x3 + x2 + x
3x4 + 6x3 - x2 + 7x - 2
We can subtract two polynomials horizontally or vertically as well.
EXAMPL E 5
Solution
Subtracting Polynomials
Find the difference: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52
Horizontal Subtraction:
13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52
= 3x4 - 4x3 + 6x2 - 1 + ( - 2x4 + 8x2 + 6x - 5)
y
Be sure to change the sign of each
term in the second polynomial.
= 13x4 - 2x4 2 + 1 - 4x3 2 + 16x2 + 8x2 2 + 6x + 1 - 1 - 52
c
Group like terms.
= x4 - 4x3 + 14x2 + 6x - 6
r
42
CHAPTER R Review
COMMENT Vertical subtraction will be
used when we divide polynomials.
■
Vertical Subtraction: We line up like terms, change the sign of each coefficient of the
second polynomial, and add.
x4
x3
x1
x2
x0
x4
3x4 - 4x3 + 6x2
- 1 =
4
2
- 3 2x
- 8x - 6x + 54 = +
x3
x2
x1
x0
3x4 - 4x3 + 6x2
- 1
4
2
- 2x
+ 8x + 6x - 5
x4 - 4x3 + 14x2 + 6x - 6
r
Which of these methods to use for adding and subtracting polynomials is up to
you. To save space, we shall most often use the horizontal format.
Now Work
PROBLEM
31
4 Multiply Polynomials
Two monomials may be multiplied using the Laws of Exponents and the
Commutative and Associative Properties. For example,
12x3 2 # 15x4 2 = 12 # 52 # 1x3 # x4 2 = 10x3 + 4 = 10x7
Products of polynomials are found by repeated use of the Distributive Property
and the Laws of Exponents. Again, you have a choice of horizontal or vertical
format.
EX AMPLE 6
Solution
Multiplying Polynomials
Find the product: 12x + 52 1x2 - x + 22
Horizontal Multiplication:
12x + 52 1x2 - x + 22 = 2x1x2 - x + 22 + 51x2 - x + 22
c
Distributive Property
= 12x # x2 - 2x # x + 2x # 22 + 15 # x2 - 5 # x + 5 # 22
c
Distributive Property
= 12x3 - 2x2 + 4x2 + 15x2 - 5x + 102
c
Law of Exponents
= 2x3 + 3x2 - x + 10
c
Combine like terms.
Vertical Multiplication: The idea here is very much like multiplying a two-digit
number by a three-digit number.
x2 -
x
2x
2x3 - 2x2 + 4x
1+2
5x2 - 5x
3
2x + 3x2 - x
Now Work
PROBLEM
+
+
47
2
5
This line is 2x(x 2 - x + 2).
+ 10
+ 10
This line is 5(x 2 - x + 2).
Sum of the above two lines
r
SECTION R.4 Polynomials
43
5 Know Formulas for Special Products
Certain products, which we call special products, occur frequently in algebra. We can
calculate them easily using the FOIL (First, Outer, Inner, Last) method of multiplying
two binomials.
Outer
First
(ax + b)(cx + d ) = ax(cx + d ) + b(cx + d )
Inner
First
Last
Outer
2
2
Inner
2
Last
2
= ax # cx + ax # d + b # cx + b # d
= acx2 + adx + bcx + bd
= acx2 + (ad + bc)x + bd
EXAMPL E 7
Using FOIL
(a) 1x - 32 1x + 32 = x2 + 3x - 3x - 9 = x2 - 9
F
(b)
(c)
(d)
(e)
O
I
L
1x + 22 = 1x + 22 1x + 22 = x + 2x + 2x + 4
1x - 32 2 = 1x - 32 1x - 32 = x2 - 3x - 3x + 9
1x + 32 1x + 12 = x2 + x + 3x + 3 = x2 + 4x +
12x + 12 13x + 42 = 6x2 + 8x + 3x + 4 = 6x2 +
2
Now Work
2
PROBLEMS
49
AND
= x2 + 4x + 4
= x2 - 6x + 9
3
11x + 4
r
57
Some products have been given special names because of their form. The
following special products are based on Examples 7(a), (b), and (c).
Difference of Two Squares
1x - a2 1x + a2 = x2 - a2
(2)
Squares of Binomials, or Perfect Squares
1x + a2 2 = x2 + 2ax + a2
(3a)
1x - a2 2 = x2 - 2ax + a2
EXAMPL E 8
(3b)
Using Special Product Formulas
(a) 1x - 52 1x + 52 = x2 - 52 = x2 - 25
(b) 1x + 72 2 = x2 + 2 # 7 # x + 72 = x2 + 14x + 49
(c) 12x + 12 2 = 12x2 2 + 2 # 1 # 2x + 12 = 4x2 + 4x + 1
(d) 13x - 42 2 = 13x2 2 - 2 # 4 # 3x + 42 = 9x2 - 24x + 16
Now Work
PROBLEMS
67, 69,
AND
Difference of two squares
Square of a binomial
Notice that we used 2x in
place of x in formula (3a).
Replace x by 3x in
formula (3b).
71
Let’s look at some more examples that lead to general formulas.
r
44
CHAPTER R Review
EX AMPLE 9
Cubing a Binomial
(a) 1x + 22 3 = 1x + 22 1x + 22 2 = 1x + 22 1x2 + 4x + 42 Formula (3a)
= 1x3 + 4x2 + 4x2 + 12x2 + 8x + 82
= x3 + 6x2 + 12x + 8
(b) 1x - 12 3 = 1x - 12 1x - 12 2 = 1x - 12 1x2 - 2x + 12 Formula (3b)
= 1x3 - 2x2 + x2 - 1x2 - 2x + 12
= x3 - 3x2 + 3x - 1
r
Cubes of Binomials, or Perfect Cubes
1x + a2 3 = x3 + 3ax2 + 3a2 x + a3
1x - a2 3 = x3 - 3ax2 + 3a2 x - a3
Now Work
EX A MPL E 1 0
PROBLEM
(4a)
(4b)
87
Forming the Difference of Two Cubes
1x - 12 1x2 + x + 12 = x1x2 + x + 12 - 11x2 + x + 12
= x3 + x2 + x - x2 - x - 1
r
= x3 - 1
EX A MPL E 11
Forming the Sum of Two Cubes
1x + 22 1x2 - 2x + 42 = x1x2 - 2x + 42 + 21x2 - 2x + 42
= x3 - 2x2 + 4x + 2x2 - 4x + 8
r
= x3 + 8
Examples 10 and 11 lead to two more special products.
Difference of Two Cubes
1x - a2 1x2 + ax + a2 2 = x3 - a3
(5)
Sum of Two Cubes
1x + a2 1x2 - ax + a2 2 = x3 + a3
(6)
6 Divide Polynomials Using Long Division
The procedure for dividing two polynomials is similar to the procedure for dividing
two integers.
SECTION R.4 Polynomials
EX AM PL E 12
45
Dividing Two Integers
Divide 842 by 15.
Solution
56
Divisor S 15 ) 842
75
92
90
2
So,
d Quotient
d Dividend
d 5 # 15 (subtract)
d 6 # 15 (subtract)
d Remainder
842
2
= 56 +
.
15
15
r
In the long-division process detailed in Example 12, the number 15 is called the
divisor, the number 842 is called the dividend, the number 56 is called the quotient,
and the number 2 is called the remainder.
To check the answer obtained in a division problem, multiply the quotient by
the divisor and add the remainder. The answer should be the dividend.
1Quotient2 1Divisor2 + Remainder = Dividend
For example, we can check the results obtained in Example 12 as follows:
1562 1152 + 2 = 840 + 2 = 842
To divide two polynomials, we first must write each polynomial in standard
form. The process then follows a pattern similar to that of Example 12. The next
example illustrates the procedure.
EX AM PL E 1 3
Dividing Two Polynomials
Find the quotient and the remainder when
3x3 + 4x2 + x + 7 is divided by x2 + 1
Solution
Each polynomial is in standard form. The dividend is 3x3 + 4x2 + x + 7, and the
divisor is x2 + 1.
NOTE Remember, a polynomial is in
standard form when its terms are
written in descending powers of x.
■
STEP 1: Divide the leading term of the dividend, 3x3, by the leading term of the
divisor, x2. Enter the result, 3x, over the term 3x3, as follows:
3x
x + 1 ) 3x3 + 4x2 + x + 7
2
STEP 2: Multiply 3x by x2 + 1, and enter the result below the dividend.
3x
x2 + 1 ) 3x3 + 4x2 + x + 7
d 3x # (x 2 + 1) = 3x 3 + 3x
3x3
+ 3x
c
Align the 3x term under the x
to make the next step easier.
STEP 3: Subtract and bring down the remaining terms.
3x
x + 1 ) 3x3 + 4x2 + x + 7
d Subtract (change the signs and add).
3x3
+ 3x
2
4x - 2x + 7 d Bring down the 4x 2 and the 7.
2
46
CHAPTER R Review
STEP 4: Repeat Steps 1–3 using 4x2 - 2x + 7 as the dividend.
3x + 4
x + 1 ) 3x3 + 4x2 + x
3x3
+ 3x
4x2 - 2x
4x2
- 2x
2
COMMENT If the degree of the divisor is
greater than the degree of the dividend,
then the process ends.
■
d
+ 7
+ 7
+ 4
+ 3
d Divide 4x 2 by x 2 to get 4.
d Multiply x 2 + 1 by 4; subtract.
Since x2 does not divide - 2x evenly (that is, the result is not a monomial), the
process ends. The quotient is 3x + 4, and the remainder is - 2x + 3.
Check: 1Quotient2 1Divisor2 + Remainder
= 13x + 42 1x2 + 12 + 1 - 2x + 32
= 3x3 + 3x + 4x2 + 4 + 1 - 2x + 32
= 3x3 + 4x2 + x + 7 = Dividend
Then
3x3 + 4x2 + x + 7
- 2x + 3
= 3x + 4 + 2
2
x + 1
x + 1
r
The next example combines the steps involved in long division.
EX A MPL E 1 4
Dividing Two Polynomials
Find the quotient and the remainder when
x4 - 3x3 + 2x - 5 is divided by x2 - x + 1
Solution
In setting up this division problem, it is necessary to leave a space for the missing x2
term in the dividend.
Divisor S
Subtract S
Subtract S
Subtract S
x2 - 2x - 3
x2 - x + 1 ) x4 - 3x3
x4 - x3 + x2
- 2x3 - x2
- 2x3 + 2x2
- 3x2
- 3x2
+ 2x - 5
+
+
+
2x
2x
4x
3x
x
d Quotient
d Dividend
- 5
- 5
- 3
- 2
d Remainder
Check: 1Quotient2 1Divisor2 + Remainder
= 1x2 - 2x - 32 1x2 - x + 12 + x - 2
= x4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2
= x4 - 3x3 + 2x - 5 = Dividend
As a result,
x - 2
x4 - 3x3 + 2x - 5
= x2 - 2x - 3 + 2
x2 - x + 1
x - x + 1
r
SECTION R.4 Polynomials
47
The process of dividing two polynomials leads to the following result:
THEOREM
Let Q be a polynomial of positive degree, and let P be a polynomial whose
degree is greater than or equal to the degree of Q. The remainder after dividing
P by Q is either the zero polynomial or a polynomial whose degree is less than
the degree of the divisor Q.
Now Work
PROBLEM
95
7 Work with Polynomials in Two Variables
A monomial in two variables x and y has the form axn ym, where a is a constant, x
and y are variables, and n and m are nonnegative integers. The degree of a monomial
is the sum of the powers of the variables.
For example,
2xy3, x2 y2, and x3 y
are all monomials that have degree 4.
A polynomial in two variables x and y is the sum of one or more monomials in
two variables. The degree of a polynomial in two variables is the highest degree of
all the monomials with nonzero coefficients.
Examples of Polynomials in Two Variables
EX AM PL E 1 5
px3 - y2
3x2 + 2x3 y + 5
Two variables,
degree is 4.
Two variables,
degree is 3.
x4 + 4x3 y - xy3 + y4
Two variables,
degree is 4.
r
Multiplying polynomials in two variables is handled in the same way as
multiplying polynomials in one variable.
Using a Special Product Formula
EX AM PL E 1 6
To multiply 12x - y2 2, use the Squares of Binomials formula (3b) with 2x instead
of x and with y instead of a.
12x - y2 2 = 12x2 2 - 2 # y # 2x + y2
= 4x2 - 4xy + y2
Now Work
PROBLEM
r
81
R.4 Assess Your Understanding
Concepts and Vocabulary
1. The polynomial 3x4 - 2x3 + 13x2 - 5 is of degree
leading coefficient is .
2. 1x2 - 42 1x2 + 42 =
3. 1x - 22 1x2 + 2x + 42 =
. The
5. Choose the degree of the monomial 3x4 y2.
(a) 3
(b) 8
(c) 6
(d) 2
6. True or False 4x -2 is a monomial of degree - 2.
.
.
4. The monomials that make up a polynomial are called which
of the following?
(a) terms (b) variables (c) factors (d) coefficients
7. True or False The degree of the product of two nonzero
polynomials equals the sum of their degrees.
8. True or False 1x + a2 1x2 + ax + a2 = x3 + a3.
48
CHAPTER R Review
Skill Building
In Problems 9–18, tell whether the expression is a monomial. If it is, name the variable(s) and the coefficient, and give the degree of the
monomial. If it is not a monomial, state why not.
9. 2x3
14. 5x2 y3
10. - 4x2
15.
11.
8x
y
8
x
16. -
2x2
y3
12. - 2x -3
13. - 2xy2
17. x2 + y2
18. 3x2 + 4
In Problems 19–28, tell whether the expression is a polynomial. If it is, give its degree. If it is not, state why not.
19. 3x2 - 5
23. 3x2 -
22. - p
21. 5
20. 1 - 4x
5
x
x2 + 5
3x3 + 2x - 1
28.
x3 - 1
x2 + x + 1
In Problems 29–48, add, subtract, or multiply, as indicated. Express your answer as a single polynomial in standard form.
24.
3
+ 2
x
25. 2y3 - 22
26. 10z2 + z
27.
29. 1x2 + 4x + 52 + 13x - 32
30. 1x3 + 3x2 + 22 + 1x2 - 4x + 42
31. 1x3 - 2x2 + 5x + 102 - 12x2 - 4x + 32
32. 1x2 - 3x - 42 - 1x3 - 3x2 + x + 52
33. 16x5 + x3 + x2 + 15x4 - x3 + 3x2 2
34. 110x5 - 8x2 2 + 13x3 - 2x2 + 62
35. 1x2 - 3x + 12 + 213x2 + x - 42
36. - 21x2 + x + 12 + 1 - 5x2 - x + 22
37. 61x3 + x2 - 32 - 412x3 - 3x2 2
38. 814x3 - 3x2 - 12 - 614x3 + 8x - 22
39. 1x2 - x + 22 + 12x2 - 3x + 52 - 1x2 + 12
40. 1x2 + 12 - 14x2 + 52 + 1x2 + x - 22
41. 91y2 - 3y + 42 - 611 - y2 2
42. 811 - y3 2 + 411 + y + y2 + y3 2
43. x1x2 + x - 42
44. 4x2 1x3 - x + 22
45. - 2x2 14x3 + 52
46. 5x3 13x - 42
47. 1x + 12 1x2 + 2x - 42
48. 12x - 32 1x2 + x + 12
In Problems 49–66, multiply the polynomials using the FOIL method. Express your answer as a single polynomial in standard form.
49. 1x + 22 1x + 42
50. 1x + 32 1x + 52
51. 12x + 52 1x + 22
52. 13x + 12 12x + 12
53. 1x - 42 1x + 22
54. 1x + 42 1x - 22
55. 1x - 32 1x - 22
56. 1x - 52 1x - 12
57. 12x + 32 1x - 22
58. 12x - 42 13x + 12
59. 1 - 2x + 32 1x - 42
60. 1 - 3x - 12 1x + 12
61. 1 - x - 22 1 - 2x - 42
62. 1 - 2x - 32 13 - x2
63. 1x - 2y2 1x + y2
64. 12x + 3y2 1x - y2
65. 1 - 2x - 3y2 13x + 2y2
66. 1x - 3y2 1 - 2x + y2
In Problems 67–90, multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard
form.
67. 1x - 72 1x + 72
68. 1x - 12 1x + 12
69. 12x + 32 12x - 32
70. 13x + 22 13x - 22
71. 1x + 42 2
72. 1x + 52 2
73. 1x - 42 2
74. 1x - 52 2
75. 13x + 42 13x - 42
76. 15x - 32 15x + 32
77. 12x - 32 2
78. 13x - 42 2
SECTION R.5 Factoring Polynomials
79. 1x + y2 1x - y2
80. 1x + 3y2 1x - 3y2
81. 13x + y2 13x - y2
82. 13x + 4y2 13x - 4y2
83. 1x + y2 2
84. 1x - y2 2
85. 1x - 2y2 2
86. 12x + 3y2 2
87. 1x - 22 3
88. 1x + 12 3
89. 12x + 12 3
90. 13x - 22 3
49
In Problems 91–106, find the quotient and the remainder. Check your work by verifying that
1Quotient2 1Divisor2 + Remainder = Dividend
91. 4x3 - 3x2 + x + 1 divided by x + 2
92. 3x3 - x2 + x - 2 divided by x + 2
93. 4x3 - 3x2 + x + 1 divided by x2
94. 3x3 - x2 + x - 2 divided by x2
95. 5x4 - 3x2 + x + 1 divided by x2 + 2
96. 5x4 - x2 + x - 2 divided by x2 + 2
97. 4x5 - 3x2 + x + 1 divided by 2x3 - 1
98. 3x5 - x2 + x - 2 divided by 3x3 - 1
99. 2x4 - 3x3 + x + 1 divided by 2x2 + x + 1
100. 3x4 - x3 + x - 2 divided by 3x2 + x + 1
101. - 4x3 + x2 - 4 divided by x - 1
102. - 3x4 - 2x - 1 divided by x - 1
103. 1 - x2 + x4 divided by x2 + x + 1
104. 1 - x2 + x4 divided by x2 - x + 1
105. x3 - a3 divided by x - a
106. x5 - a5 divided by x - a
Explaining Concepts: Discussion and Writing
107. Explain why the degree of the product of two nonzero
polynomials equals the sum of their degrees.
108. Explain why the degree of the sum of two polynomials of
different degrees equals the larger of their degrees.
109. Give a careful statement about the degree of the sum of two
polynomials of the same degree.
110. Do you prefer adding two polynomials using the horizontal
method or the vertical method? Write a brief position paper
defending your choice.
111. Do you prefer to memorize the rule for the square of a
binomial 1x + a2 2 or to use FOIL to obtain the product?
Write a brief position paper defending your choice.
R.5 Factoring Polynomials
OBJECTIVES 1 Factor the Difference of Two Squares and the Sum and Difference of Two
Cubes (p. 50)
2 Factor Perfect Squares (p. 51)
3 Factor a Second-Degree Polynomial: x2 + Bx + C (p. 52)
4 Factor by Grouping (p. 53)
5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A ≠ 1 (p. 54)
6 Complete the Square (p. 56)
Consider the following product:
12x + 32 1x - 42 = 2x2 - 5x - 12
The two polynomials on the left side are called factors of the polynomial on the
right side. Expressing a given polynomial as a product of other polynomials—that
is, finding the factors of a polynomial—is called factoring.
50
CHAPTER R Review
We shall restrict our discussion here to factoring polynomials in one variable
into products of polynomials in one variable, where all coefficients are integers. We
call this factoring over the integers.
Any polynomial can be written as the product of 1 times itself or as - 1 times
its additive inverse. If a polynomial cannot be written as the product of two other
polynomials (excluding 1 and - 1), then the polynomial is prime. When a polynomial
has been written as a product consisting only of prime factors, it is factored completely.
Examples of prime polynomials (over the integers) are
COMMENT Over the real numbers,
3x + 4 factors into 3 1x + 43 2 . It is
the noninteger 43 that causes 3x + 4
to be prime over the integers. In most
instances, we will be factoring over the
integers.
■
EX AMPLE 1
2, 3, 5, x, x + 1, x - 1, 3x + 4, x2 + 4
The first factor to look for in a factoring problem is a common monomial
factor present in each term of the polynomial. If one is present, use the Distributive
Property to factor it out. Continue factoring out monomial factors until none are
left.
Identifying Common Monomial Factors
Polynomial
Common
Monomial
Factor
Remaining
Factor
Factored Form
2x + 4
2
x + 2
2x + 4 = 21x + 22
3x - 6
3
x - 2
3x - 6 = 31x - 22
2x - 4x + 8
2
x - 2x + 4
2x2 - 4x + 8 = 21x2 - 2x + 42
8x - 12
4
2x - 3
8x - 12 = 412x - 32
x + x
x
x + 1
x2 + x = x1x + 12
x3 - 3x2
x2
x - 3
x3 - 3x2 = x2 1x - 32
6x2 + 9x
3x
2x + 3
6x2 + 9x = 3x12x + 32
2
2
2
r
Notice that once all common monomial factors have been removed from a
polynomial, the remaining factor is either a prime polynomial of degree 1 or a
polynomial of degree 2 or higher. (Do you see why?)
Now Work
PROBLEM
9
1 Factor the Difference of Two Squares and the Sum
and Difference of Two Cubes
When you factor a polynomial, first check for common monomial factors. Then see
whether you can use one of the special formulas discussed in the previous section.
Difference of Two Squares
Perfect Squares
Sum of Two Cubes
Difference of Two Cubes
EX AMPLE 2
x2
x2
x2
x3
x3
+
+
-
a2 = 1x - a2 1x + a2
2ax + a2 = 1x + a2 2
2ax + a2 = 1x - a2 2
a3 = 1x + a2 1x2 - ax + a2 2
a3 = 1x - a2 1x2 + ax + a2 2
Factoring the Difference of Two Squares
Factor completely: x2 - 4
Solution
Note that x2 - 4 is the difference of two squares, x2 and 22.
x2 - 4 = 1x - 22 1x + 22
r
SECTION R.5 Factoring Polynomials
EXAMPL E 3
51
Factoring the Difference of Two Cubes
Factor completely: x3 - 1
Solution
EXAMPL E 4
Because x3 - 1 is the difference of two cubes, x3 and 13,
x3 - 1 = 1x - 12 1x2 + x + 12
r
Factoring the Sum of Two Cubes
Factor completely: x3 + 8
Solution
EXAMPL E 5
Because x3 + 8 is the sum of two cubes, x3 and 23,
x3 + 8 = 1x + 22 1x2 - 2x + 42
r
Factoring the Difference of Two Squares
Factor completely: x4 - 16
Solution
Because x4 - 16 is the difference of two squares, x4 = 1x2 2 and 16 = 42,
2
x4 - 16 = 1x2 - 42 1x2 + 42
But x2 - 4 is also the difference of two squares. Then,
x4 - 16 = 1x2 - 42 1x2 + 42 = 1x - 22 1x + 22 1x2 + 42
Now Work
PROBLEMS
19
AND
r
37
2 Factor Perfect Squares
When the first term and third term of a trinomial are both positive and are perfect
squares, such as x2, 9x2, 1, and 4, check to see whether the trinomial is a perfect square.
EXAMPL E 6
Factoring a Perfect Square
Factor completely: x2 + 6x + 9
Solution
The first term, x2, and the third term, 9 = 32, are perfect squares. Because the middle
term, 6x, is twice the product of x and 3, we have a perfect square.
x2 + 6x + 9 = 1x + 32 2
EXAMPL E 7
r
Factoring a Perfect Square
Factor completely: 9x2 - 6x + 1
Solution
The first term, 9x2 = 13x2 2, and the third term, 1 = 12, are perfect squares. Because
the middle term, - 6x, is - 2 times the product of 3x and 1, we have a perfect square.
9x2 - 6x + 1 = 13x - 12 2
EXAMPL E 8
r
Factoring a Perfect Square
Factor completely: 25x2 + 30x + 9
Solution
The first term, 25x2 = 15x2 2, and the third term, 9 = 32, are perfect squares.
Because the middle term, 30x, is twice the product of 5x and 3, we have a perfect square.
25x2 + 30x + 9 = 15x + 32 2
Now Work
PROBLEMS
29
AND
103
r
52
CHAPTER R Review
If a trinomial is not a perfect square, it may be possible to factor it using the
technique discussed next.
3 Factor a Second-Degree Polynomial: x2 + Bx + C
The idea behind factoring a second-degree polynomial like x2 + Bx + C is to see
whether it can be made equal to the product of two (possibly equal) first-degree
polynomials.
For example, consider
1x + 32 1x + 42 = x2 + 7x + 12
The factors of x2 + 7x + 12 are x + 3 and x + 4. Notice the following:
x2 + 7x + 12 = (x + 3)(x + 4)
12 is the product of 3 and 4
7 is the sum of 3 and 4
In general, if x2 + Bx + C = 1x + a2 1x + b2 = x2 + (a + b)x + ab, then
ab = C and a + b = B.
To factor a second-degree polynomial x2 + Bx + C, find integers whose
product is C and whose sum is B. That is, if there are numbers a, b, where
ab = C and a + b = B, then
x2 + Bx + C = 1x + a2 1x + b2
EX AMPLE 9
Factoring a Trinomial
Factor completely: x2 + 7x + 10
Solution
First determine all pairs of integers whose product is 10, and then compute their sums.
Integers whose product is 10
1, 10
- 1, - 10
2, 5
- 2, - 5
Sum
11
- 11
7
-7
The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the
middle term. As a result,
x2 + 7x + 10 = 1x + 22 1x + 52
EX A MPL E 1 0
r
Factoring a Trinomial
Factor completely: x2 - 6x + 8
Solution
First determine all pairs of integers whose product is 8, and then compute each sum.
Integers whose product is 8
1, 8
- 1, - 8
2, 4
- 2, - 4
Sum
9
-9
6
-6
Since - 6 is the coefficient of the middle term,
x2 - 6x + 8 = 1x - 22 1x - 42
r
SECTION R.5 Factoring Polynomials
EX AM PL E 11
53
Factoring a Trinomial
Factor completely: x2 - x - 12
Solution
First determine all pairs of integers whose product is - 12, and then compute each sum.
Integers whose product is − 12
1, - 12
- 1, 12
2, - 6
- 2, 6
3, - 4
- 3, 4
Sum
- 11
11
-4
4
-1
1
Since - 1 is the coefficient of the middle term,
x2 - x - 12 = 1x + 32 1x - 42
EX AM PL E 1 2
r
Factoring a Trinomial
Factor completely: x2 + 4x - 12
Solution
The integers - 2 and 6 have a product of - 12 and have the sum 4. So,
x2 + 4x - 12 = 1x - 22 1x + 62
r
To avoid errors in factoring, always check your answer by multiplying it out to
see whether the result equals the original expression.
When none of the possibilities works, the polynomial is prime.
EX AM PL E 1 3
Identifying a Prime Polynomial
Show that x2 + 9 is prime.
Solution
First list the pairs of integers whose product is 9, and then compute their sums.
Integers whose product is 9
1, 9
- 1, - 9
3, 3
- 3, - 3
Sum
10
- 10
6
-6
Since the coefficient of the middle term in x2 + 9 = x2 + 0x + 9 is 0 and none of
the sums equals 0, we conclude that x2 + 9 is prime.
r
Example 13 demonstrates a more general result:
THEOREM
Any polynomial of the form x2 + a2, a real, is prime.
Now Work
PROBLEMS
43
AND
87
4 Factor by Grouping
Sometimes a common factor does not occur in every term of the polynomial but
does occur in each of several groups of terms that together make up the polynomial.
When this happens, the common factor can be factored out of each group by means
of the Distributive Property. This technique is called factoring by grouping.
EX AM PL E 1 4
Solution
Factoring by Grouping
Factor completely by grouping: 1x2 + 22x + 1x2 + 22 # 3
Notice the common factor x2 + 2. Applying the Distributive Property yields
1x2 + 22x + 1x2 + 22 # 3 = 1x2 + 22 1x + 32
Since x2 + 2 and x + 3 are prime, the factorization is complete.
r
54
CHAPTER R Review
The next example shows a factoring problem that occurs in calculus.
EX A MPL E 1 5
Solution
Factoring by Grouping
Factor completely by grouping: 31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3
Here, 1x - 12 2 1x + 22 3 is a common factor of both 31x - 12 2 1x + 22 4 and
41x - 12 3 1x + 22 3. As a result,
31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3 = 1x - 12 2 1x + 22 3 3 31x + 22 + 41x - 12 4
= 1x - 12 2 1x + 22 3 3 3x + 6 + 4x - 44
= 1x - 12 2 1x + 22 3 17x + 22
EX A MPL E 1 6
r
Factoring by Grouping
Factor completely by grouping: x3 - 4x2 + 2x - 8
Solution
To see whether factoring by grouping will work, group the first two terms and the
last two terms. Then look for a common factor in each group. In this example, factor
x2 from x3 - 4x2 and 2 from 2x - 8. The remaining factor in each case is the same,
x - 4. This means that factoring by grouping will work, as follows:
x3 - 4x2 + 2x - 8 = 1x3 - 4x2 2 + 12x - 82
= x2 1x - 42 + 21x - 42
= 1x - 42 1x2 + 22
Since x2 + 2 and x - 4 are prime, the factorization is complete.
Now Work
PROBLEMS
55
AND
r
131
5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A 3 1
To factor a second-degree polynomial Ax2 + Bx + C, when A ≠ 1 and A, B, and
C have no common factors, follow these steps:
Steps for Factoring Ax2 + Bx + C,
When A 3 1 and A, B, and C Have No Common Factors
STEP 1:
STEP 2:
STEP 3:
STEP 4:
EX AMPLE 17
Find the value of AC.
Find a pair of integers whose product is AC and that add up to B. That
is, find a and b such that ab = AC and a + b = B.
Write Ax2 + Bx + C = Ax2 + ax + bx + C.
Factor this last expression by grouping.
Factoring a Trinomial
Factor completely: 2x2 + 5x + 3
Solution
Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, we find that A = 2, B = 5, and C = 3.
STEP 1: The value of AC is 2 # 3 = 6.
STEP 2: Determine the pairs of integers whose product is AC = 6 and compute
their sums.
Integers whose product is 6
1, 6
- 1, - 6
2, 3
- 2, - 3
Sum
7
-7
5
-5
SECTION R.5 Factoring Polynomials
55
STEP 3: The integers whose product is 6 that add up to B = 5 are 2 and 3.
2x2 + 5x + 3 = 2x2 + 2x + 3x + 3
STEP 4: Factor by grouping.
2x2 + 2x + 3x + 3 = 12x2 + 2x2 + 13x + 32
= 2x1x + 12 + 31x + 12
= 1x + 12 12x + 32
As a result,
EX AMPL E 1 8
2x2 + 5x + 3 = 1x + 12 12x + 32
r
Factoring a Trinomial
Factor completely: 2x2 - x - 6
Solution
Comparing 2x2 - x - 6 to Ax2 + Bx + C, we find that A = 2, B = - 1, and
C = - 6.
STEP 1: The value of AC is 2 # 1 - 62 = - 12.
STEP 2: Determine the pairs of integers whose product is AC = - 12 and compute
their sums.
Integers whose product is − 12
1, - 12
- 1, 12
2, - 6
- 2, 6
3, - 4
- 3, 4
Sum
- 11
11
-4
4
-1
1
STEP 3: The integers whose product is - 12 that add up to B = - 1 are - 4 and 3.
2x2 - x - 6 = 2x2 - 4x + 3x - 6
STEP 4: Factor by grouping.
2x2 - 4x + 3x - 6 = 12x2 - 4x2 + 13x - 62
= 2x1x - 22 + 31x - 22
= 1x - 22 12x + 32
As a result,
Now Work
2x2 - x - 6 = 1x - 22 12x + 32
PROBLEM
61
SUMMARY
Type of Polynomial
Method
Example
Any polynomial
Look for common monomial factors.
(Always do this first!)
6x2 + 9x = 3x12x + 32
Binomials of
degree 2 or higher
Check for a special product:
Difference of two squares, x2 - a2
Difference of two cubes, x3 - a3
Sum of two cubes, x3 + a3
x2 - 16 = 1x - 42 1x + 42
x3 - 64 = 1x - 42 1x2 + 4x + 162
x3 + 27 = 1x + 32 1x2 - 3x + 92
Trinomials of degree 2
Four or more terms
Check for a perfect square, 1x { a2 2
Factoring x2 + Bx + C (p. 52)
Factoring Ax2 + Bx + C (p. 54)
x2 + 8x + 16 = 1x + 42 2
x2 - 10x + 25 = 1x - 52 2
x2 - x - 2 = 1x - 22 1x + 12
6x2 + x - 1 = 12x + 12 13x - 12
Grouping
2x3 - 3x2 + 4x - 6 = 12x - 32 1x2 + 22
r
56
CHAPTER R Review
6 Complete the Square
The idea behind completing the square in one variable is to “adjust” an expression
of the form x2 + bx to make it a perfect square. Perfect squares are trinomials of
the form
x2 + 2ax + a2 = (x + a)2 or x2 - 2ax + a2 = (x - a)2
For example, x2 + 6x + 9 is a perfect square because x2 + 6x + 9 = (x + 3)2. And
p2 - 12p + 36 is a perfect square because p2 - 12p + 36 = (p - 6)2.
So how do we “adjust” x2 + bx to make it a perfect square? We do it by adding
a number. For example, to make x2 + 6x a perfect square, add 9. But how do we
know to add 9? If we divide the coefficient of the first-degree term, 6, by 2, and then
square the result, we obtain 9. This approach works in general.
1 2
WARNING To use a bb to complete
2
the square, the coefficient of the x 2
term must be 1.
■
EX A MPL E 1 9
Completing the Square of x 2 + bx
1
2
and then square the result. That is, determine the value of b in x2 + bx and
1 2
compute a b b .
2
Identify the coefficient of the first-degree term. Multiply this coefficient by
Completing the Square
Determine the number that must be added to each expression to complete the
square. Then factor the expression.
Start
Add
Result
Factored Form
y2 + 8y
a
2
1#
8b = 16
2
y2 + 8y + 16
(y + 4)2
x2 + 12x
a
2
1#
12b = 36
2
x2 + 12x + 36
(x + 6)2
a2 - 20a
a
2
1#
( - 20) b = 100
2
a2 - 20a + 100
(a - 10)2
p2 - 5p
a
2
1#
25
( - 5) b =
2
4
p2 - 5p +
25
4
ap -
5 2
b
2
r
Notice that the factored form of a perfect square is either
y
y
4
Area 5 y 2
Area 5 4y
Figure 27
4
Area
5 4y
b 2
b 2
b 2
b 2
x2 + bx + a b = ax + b or x2 - bx + a b = ax - b
2
2
2
2
Now Work
PROBLEM
73
Are you wondering why we refer to making an expression a perfect square
as “completing the square”? Look at the square in Figure 27. Its area is (y + 4)2.
The yellow area is y2 and each orange area is 4y (for a total area of 8y). The sum of
these areas is y2 + 8y. To complete the square, we need to add the area of the green
region: 4 # 4 = 16. As a result, y2 + 8y + 16 = (y + 4)2.
SECTION R.5 Factoring Polynomials
57
R.5 Assess Your Understanding
Concepts and Vocabulary
1. If factored completely, 3x3 - 12x =
4. Choose the best description of x2 - 64.
(a) Prime
(b) Difference of two squares
(c) Difference of two cubes
(d) Perfect Square
.
2. If a polynomial cannot be written as the product of two
other polynomials (excluding 1 and - 1), then the polynomial
is said to be
.
5. Choose the complete factorization of 4x2 - 8x - 60.
(a) 21x + 32 1x - 52
(b) 41x2 - 2x - 152
(c) 12x + 62 12x - 102 (d) 41x + 32 1x - 52
3. For x + Bx + C = 1x + a2 1x + b2 , which of the
following must be true?
(a) ab = B and a + b = C
(b) a + b = C and a - b = B
(c) ab = C and a + b = B
(d) ab = B and a - b = C
2
6. To complete the square of x2 + bx, use which of the following?
1 2
1
(a) 12b2 2 (b) 2b2 (c) a bb (d) b2
2
2
7. True or False The polynomial x2 + 4 is prime.
8. True or False 3x3 - 2x2 - 6x + 4 = 13x - 22 1x2 + 22.
Skill Building
In Problems 9–18, factor each polynomial by removing the common monomial factor.
9. 3x + 6
11. ax2 + a
12. ax - a
10. 7x - 14
15. 2x2 - 2x
14. x3 - x2 + x
16. 3x2 - 3x
13. x3 + x2 + x
17. 3x2 y - 6xy2 + 12xy
18. 60x2 y - 48xy2 + 72x3 y
In Problems 19–26, factor the difference of two squares.
19. x2 - 1
20. x2 - 4
21. 4x2 - 1
22. 9x2 - 1
23. x2 - 16
24. x2 - 25
25. 25x2 - 4
26. 36x2 - 9
In Problems 27–36, factor the perfect squares.
27. x2 + 2x + 1
28. x2 - 4x + 4
29. x2 + 4x + 4
30. x2 - 2x + 1
31. x2 - 10x + 25
32. x2 + 10x + 25
33. 4x2 + 4x + 1
34. 9x2 + 6x + 1
35. 16x2 + 8x + 1
36. 25x2 + 10x + 1
In Problems 37–42, factor the sum or difference of two cubes.
37. x3 - 27
38. x3 + 125
39. x3 + 27
40. 27 - 8x3
41. 8x3 + 27
42. 64 - 27x3
In Problems 43–54, factor each polynomial.
43. x2 + 5x + 6
44. x2 + 6x + 8
45. x2 + 7x + 6
46. x2 + 9x + 8
47. x2 + 7x + 10
48. x2 + 11x + 10
49. x2 - 10x + 16
50. x2 - 17x + 16
51. x2 - 7x - 8
52. x2 - 2x - 8
53. x2 + 7x - 8
54. x2 + 2x - 8
In Problems 55–60, factor by grouping.
55. 2x2 + 4x + 3x + 6
56. 3x2 - 3x + 2x - 2
57. 2x2 - 4x + x - 2
58. 3x2 + 6x - x - 2
59. 6x2 + 9x + 4x + 6
60. 9x2 - 6x + 3x - 2
In Problems 61–72, factor each polynomial.
61. 3x2 + 4x + 1
62. 2x2 + 3x + 1
63. 2z2 + 5z + 3
64. 6z2 + 5z + 1
65. 3x2 + 2x - 8
66. 3x2 + 10x + 8
67. 3x2 - 2x - 8
68. 3x2 - 10x + 8
69. 3x2 + 14x + 8
70. 3x2 - 14x + 8
71. 3x2 + 10x - 8
72. 3x2 - 10x - 8
In Problems 73–78, determine what number should be added to complete the square of each expression. Then factor each expression.
73. x2 + 10x
74. p2 + 14p
76. x2 - 4x
77. x2 -
1
x
2
75. y2 - 6y
78. x2 +
1
x
3
58
CHAPTER R Review
Mixed Practice
In Problems 79–126, factor each polynomial completely. If the polynomial cannot be factored, say it is prime.
79. x2 - 36
80. x2 - 9
81. 2 - 8x2
82. 3 - 27x2
83. x2 + 11x + 10
84. x2 + 5x + 4
85. x2 - 10x + 21
86. x2 - 6x + 8
87. 4x2 - 8x + 32
88. 3x2 - 12x + 15
89. x2 + 4x + 16
90. x2 + 12x + 36
91. 15 + 2x - x2
92. 14 + 6x - x2
93. 3x2 - 12x - 36
94. x3 + 8x2 - 20x
95. y4 + 11y3 + 30y2
96. 3y3 - 18y2 - 48y
97. 4x2 + 12x + 9
98. 9x2 - 12x + 4
99. 6x2 + 8x + 2
100. 8x2 + 6x - 2
101. x4 - 81
102. x4 - 1
103. x6 - 2x3 + 1
104. x6 + 2x3 + 1
105. x7 - x5
106. x8 - x5
107. 16x2 + 24x + 9
108. 9x2 - 24x + 16
109. 5 + 16x - 16x2
110. 5 + 11x - 16x2
111. 4y2 - 16y + 15
112. 9y2 + 9y - 4
113. 1 - 8x2 - 9x4
114. 4 - 14x2 - 8x4
115. x1x + 32 - 61x + 32
116. 513x - 72 + x13x - 72
117. 1x + 22 2 - 51x + 22
118. 1x - 12 2 - 21x - 12
119. 13x - 22 3 - 27
120. 15x + 12 3 - 1
121. 31x2 + 10x + 252 - 41x + 52
122. 71x2 - 6x + 92 + 51x - 32
123. x3 + 2x2 - x - 2
124. x3 - 3x2 - x + 3
125. x4 - x3 + x - 1
126. x4 + x3 + x + 1
Applications and Extensions
In Problems 127–136, expressions that occur in calculus are given. Factor each expression completely.
127. 213x + 42 2 + 12x + 32 # 213x + 42 # 3
129. 2x12x + 52 + x2 # 2
128. 512x + 12 2 + 15x - 62 # 212x + 12 # 2
130. 3x2 18x - 32 + x3 # 8
131. 21x + 32 1x - 22 3 + 1x + 32 2 # 31x - 22 2
132. 41x + 52 3 1x - 12 2 + 1x + 52 4 # 21x - 12
135. 213x - 52 # 312x + 12 3 + 13x - 52 2 # 312x + 12 2 # 2
136. 314x + 52 2 # 415x + 12 2 + 14x + 52 3 # 215x + 12 # 5
133. 14x - 32 2 + x # 214x - 32 # 4
137. Show that x2 + 4 is prime.
134. 3x2 13x + 42 2 + x3 # 213x + 42 # 3
138. Show that x2 + x + 1 is prime.
Explaining Concepts: Discussion and Writing
139. Make up a polynomial that factors into a perfect square.
140. Explain to a fellow student what you look for first when
presented with a factoring problem. What do you do next?
R.6 Synthetic Division
OBJECTIVE 1 Divide Polynomials Using Synthetic Division (p. 58)
1 Divide Polynomials Using Synthetic Division
To find the quotient as well as the remainder when a polynomial of degree 1 or
higher is divided by x - c, a shortened version of long division, called synthetic
division, makes the task simpler.
SECTION R.6 Synthetic Division
59
To see how synthetic division works, first consider long division for dividing the
polynomial 2x3 - x2 + 3 by x - 3.
d Quotient
2x2 + 5x + 15
3
2
x - 3 ) 2x - x
+ 3
2x3 - 6x2
5x2
5x2 - 15x
15x + 3
15x - 45
48 d Remainder
Check: 1Divisor2 # 1Quotient2 + Remainder
= 1x - 32 12x2 + 5x + 152 + 48
= 2x3 + 5x2 + 15x - 6x2 - 15x - 45 + 48
= 2x3 - x2 + 3
The process of synthetic division arises from rewriting the long division in a
more compact form, using simpler notation. For example, in the long division above,
the terms in blue are not really necessary because they are identical to the terms
directly above them. With these terms removed, we have
2x2 + 5x + 15
x - 3 ) 2x3 - x2
+ 3
2
- 6x
5x2
- 15x
15x
- 45
48
Most of the x’s that appear in this process can also be removed, provided that we
are careful about positioning each coefficient. In this regard, we will need to use 0 as
the coefficient of x in the dividend, because that power of x is missing. Now we have
2x2 + 5x + 15
x - 3) 2 - 1
0
- 6
5
- 15
15
3
- 45
48
We can make this display more compact by moving the lines up until the numbers
in blue align horizontally.
2x2 + 5x + 15
0
3
x - 3) 2 - 1
- 6 - 15 - 45
~
5
15
48
Row 1
Row 2
Row 3
Row 4
Because the leading coefficient of the divisor is always 1, the leading coefficient
of the dividend will also be the leading coefficient of the quotient. So we place the
leading coefficient of the quotient, 2, in the circled position. Now, the first three
numbers in row 4 are precisely the coefficients of the quotient, and the last number
CHAPTER R Review
in row 4 is the remainder. Since row 1 is not really needed, we can compress the
process to three rows, where the bottom row contains both the coefficients of the
quotient and the remainder.
x - 3) 2
- 1
0
3 Row 1
- 6 - 15 - 45 Row 2 (subtract)
5
15
48 Row 3
2
3
45
48
S3
0
15
15
*
S3
*
2
- 1
6
5
*
x - 3) 2
S3
Recall that the entries in row 3 are obtained by subtracting the entries in row 2
from those in row 1. Rather than subtracting the entries in row 2, we can change the
sign of each entry and add. With this modification, our display will look like this:
Row 1
Row 2 (add)
Row 3
Notice that the entries in row 2 are three times the prior entries in row 3. Our
last modification to the display replaces the x - 3 by 3. The entries in row 3 give the
quotient and the remainder, as shown next.
3) 2
0
15
15
-1
6
5
3
45
48
Row 1
Row 2 (add)
Row 3
c
e
2
Quotient
Remainder
c
e
2
2x + 5x + 15
48
Let’s go through an example step by step.
EX AMPLE 1
Using Synthetic Division to Find the Quotient and Remainder
Use synthetic division to find the quotient and remainder when
x3 - 4x2 - 5 is divided by x - 3
STEP 1: Write the dividend in descending powers of x. Then copy the coefficients,
remembering to insert a 0 for any missing powers of x.
1
-4 0
-5
Row 1
STEP 2: Insert the usual division symbol. In synthetic division, the divisor is of the
form x - c, and c is the number placed to the left of the division symbol.
Here, since the divisor is x - 3, insert 3 to the left of the division symbol.
3) 1
-4 0
-5
Row 1
STEP 3: Bring the 1 down two rows, and enter it in row 3.
3) 1 - 4 0 - 5
T
1
STEP 4: Multiply the latest entry in row 3 by
column over to the right.
3) 1
-4 0
3
Row 1
Row 2
Row 3
3, and place the result in row 2, one
- 5 Row 1
Row 2
Row 3
*
S3
1
STEP 5: Add the entry in row 2 to the entry above it in row 1, and enter the sum in
row 3.
3 ) 1 - 4 0 - 5 Row 1
Row 2
3
Row 3
1 -1
S3
Solution
*
60
SECTION R.6 Synthetic Division
61
STEP 6: Repeat Steps 4 and 5 until no more entries are available in row 1.
3) 1
S3
*
S3
*
*
S3
-4
0 - 5 Row 1
3 - 3 - 9 Row 2
1 - 1 - 3 - 14 Row 3
STEP 7: The final entry in row 3, the - 14, is the remainder; the other entries in row 3,
the 1, - 1, and - 3, are the coefficients (in descending order) of a polynomial whose
degree is 1 less than that of the dividend. This is the quotient. That is,
Quotient = x2 - x - 3
Remainder = - 14
Check: 1Divisor2 1Quotient2 + Remainder
= 1x - 32 1x2 - x - 32 + 1 - 142
= 1x3 - x2 - 3x - 3x2 + 3x + 92 + 1 - 142
r
= x3 - 4x2 - 5 = Dividend
Let’s do an example in which all seven steps are combined.
EXAMPL E 2
Using Synthetic Division to Verify a Factor
Use synthetic division to show that x + 3 is a factor of
2x5 + 5x4 - 2x3 + 2x2 - 2x + 3
Solution
The divisor is x + 3 = x - ( - 3), so place - 3 to the left of the division symbol. Then
the row 3 entries will be multiplied by - 3, entered in row 2, and added to row 1.
- 3) 2
2
5
-6
-1
-2
3
1
2
-3
-1
-2
3
1
3
-3
0
Row 1
Row 2
Row 3
Because the remainder is 0, we have
1Divisor2 1Quotient2 + Remainder
= 1x + 32 12x4 - x3 + x2 - x + 12 = 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3
r
As we see, x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3.
As Example 2 illustrates, the remainder after division gives information about
whether the divisor is, or is not, a factor. We shall have more to say about this in Chapter 5.
Now Work
PROBLEMS
9
AND
19
R.6 Assess Your Understanding
Concepts and Vocabulary
1. To check division, the expression that is being divided, the dividend, should equal the product of the
the
.
and the
2. To divide 2x3 - 5x + 1 by x + 3 using synthetic division, the first step is to write
.
)
3. Choose the division problem that cannot be done using synthetic division.
(b) x4 - 3 is divided by x + 1
(a) 2x3 - 4x2 + 6x - 8 is divided by x - 8
5
2
(c) x + 3x - 9x + 2 is divided by x + 10
(d) x4 - 5x3 + 3x2 - 9x + 13 is divided by
4. Choose the correct conclusion based on the following synthetic division: - 5 ) 2
3 - 38 - 15
x2 + 5
- 10 35
15
2
-7 - 3
0
(b) x - 5 is a factor of
2x3 + 3x2 - 38x - 15
(a) x + 5 is a factor of
2x3 + 3x2 - 38x - 15
(c) x + 5 is not a factor of 2x3 + 3x2 - 38x - 15
(d) x - 5 is not a factor of 2x3 + 3x2 - 38x - 15
5. True or False In using synthetic division, the divisor is always a polynomial of degree 1, whose leading coefficient is 1.
6. True or False - 2 ) 5
3
2
1
5x3 + 3x2 + 2x + 1
- 31
= 5x2 - 7x + 16 +
.
- 10 14 - 32 means
x + 2
x + 2
5
- 7 16 - 31
plus
62
CHAPTER R Review
Skill Building
In Problems 7–18, use synthetic division to find the quotient and remainder when:
7. x3 - x2 + 2x + 4 is divided by x - 2
8. x3 + 2x2 - 3x + 1 is divided by x + 1
9. 3x3 + 2x2 - x + 3 is divided by x - 3
10. - 4x3 + 2x2 - x + 1 is divided by x + 2
11. x5 - 4x3 + x is divided by x + 3
12. x4 + x2 + 2 is divided by x - 2
13. 4x6 - 3x4 + x2 + 5 is divided by x - 1
14. x5 + 5x3 - 10 is divided by x + 1
15. 0.1x3 + 0.2x is divided by x + 1.1
16. 0.1x2 - 0.2 is divided by x + 2.1
17. x5 - 1 is divided by x - 1
18. x5 + 1 is divided by x + 1
In Problems 19–28, use synthetic division to determine whether x - c is a factor of the given polynomial.
19. 4x3 - 3x2 - 8x + 4; x - 2
20. - 4x3 + 5x2 + 8; x + 3
21. 3x4 - 6x3 - 5x + 10; x - 2
22. 4x4 - 15x2 - 4; x - 2
23. 3x6 + 82x3 + 27; x + 3
24. 2x6 - 18x4 + x2 - 9; x + 3
25. 4x6 - 64x4 + x2 - 15; x + 4
1
27. 2x4 - x3 + 2x - 1; x 2
26. x6 - 16x4 + x2 - 16; x + 4
1
28. 3x4 + x3 - 3x + 1; x +
3
Applications and Extensions
29. Find the sum of a, b, c, and d if
x3 - 2x2 + 3x + 5
d
= ax2 + bx + c +
x + 2
x + 2
Explaining Concepts: Discussion and Writing
30. When dividing a polynomial by x - c, do you prefer to use long division or synthetic division? Does the value of c make a
difference to you in choosing? Give reasons.
R.7 Rational Expressions
OBJECTIVES 1 Reduce a Rational Expression to Lowest Terms (p. 62)
2 Multiply and Divide Rational Expressions (p. 63)
3 Add and Subtract Rational Expressions (p. 64)
4 Use the Least Common Multiple Method (p. 66)
5 Simplify Complex Rational Expressions (p. 68)
1 Reduce a Rational Expression to Lowest Terms
If we form the quotient of two polynomials, the result is called a rational expression.
Some examples of rational expressions are
(a)
x3 + 1
x
(b)
3x2 + x - 2
x2 + 5
(c)
x
2
x - 1
(d)
xy2
1x - y2 2
Expressions (a), (b), and (c) are rational expressions in one variable, x, whereas
(d) is a rational expression in two variables, x and y.
Rational expressions are described in the same manner as rational numbers. In
expression (a), the polynomial x3 + 1 is the numerator, and x is the denominator.
When the numerator and denominator of a rational expression contain no common
factors (except 1 and - 1), we say that the rational expression is reduced to lowest
terms, or simplified.
The polynomial in the denominator of a rational expression cannot be equal
x3 + 1
to 0 because division by 0 is not defined. For example, for the expression
,x
x
cannot take on the value 0. The domain of the variable x is 5 x 0 x ≠ 06 .
SECTION R.7 Rational Expressions
63
A rational expression is reduced to lowest terms by factoring the numerator
and the denominator completely and canceling any common factors using the
Cancellation Property:
ac
a
=
bc
b
EXAMPL E 1
(1)
Reducing a Rational Expression to Lowest Terms
Reduce to lowest terms:
Solution
if b ≠ 0, c ≠ 0
x2 + 4x + 4
x2 + 3x + 2
Begin by factoring the numerator and the denominator.
x2 + 4x + 4 = 1x + 22 1x + 22
x2 + 3x + 2 = 1x + 22 1x + 12
WARNING Apply the Cancellation
Property only to rational expressions
written in factored form. Be sure to
cancel only common factors, not
common terms!
■
EXAMPL E 2
Since a common factor, x + 2, appears, the original expression is not in lowest terms.
To reduce it to lowest terms, use the Cancellation Property:
1x + 22 1x + 22
x + 2
x2 + 4x + 4
=
=
2
1x + 22 1x + 12
x + 1
x + 3x + 2
x ≠ - 2, - 1
r
Reducing Rational Expressions to Lowest Terms
Reduce each rational expression to lowest terms.
(a)
Solution
(a)
(b)
x3 - 8
x3 - 2x2
(b)
8 - 2x
x - x - 12
2
1x - 22 1x2 + 2x + 42
x2 + 2x + 4
x3 - 8
=
=
x3 - 2x2
x2 1x - 22
x2
x ≠ 0, 2
21 - 12 1x - 42
214 - x2
8 - 2x
-2
=
=
=
1x
42
1x
+
32
1x
42
1x
+
32
x
+ 3
x - x - 12
2
Now Work
PROBLEM
x ≠ - 3, 4
r
7
2 Multiply and Divide Rational Expressions
The rules for multiplying and dividing rational expressions are the same as the rules
a
c
for multiplying and dividing rational numbers. If and , b ≠ 0, d ≠ 0, are two
b
d
rational expressions, then
a#c
ac
=
b d
bd
a
b
a d
ad
= # =
c
b c
bc
d
if b ≠ 0, d ≠ 0
if b ≠ 0, c ≠ 0, d ≠ 0
(2)
(3)
In using equations (2) and (3) with rational expressions, be sure first to factor
each polynomial completely so that common factors can be canceled. Leave your
answer in factored form.
64
CHAPTER R Review
EX AMPLE 3
Multiplying and Dividing Rational Expressions
Perform the indicated operation and simplify the result. Leave your answer in
factored form.
x + 3
2
2
x - 2x + 1 # 4x + 4
x2 - 4
(b) 2
(a)
3
2
x + x
x + x - 2
x - x - 12
x3 - 8
Solution
(a)
1x - 12 2
41x2 + 12
x2 - 2x + 1 # 4x2 + 4
#
=
x3 + x
x2 + x - 2
x1x2 + 12 1x + 22 1x - 12
1x - 12 2 142 1x2 + 12
=
x 1x2 + 12 1x + 22 1x - 12
41x - 12
=
x ≠ - 2, 0, 1
x1x + 22
x + 3
x + 3 #
x3 - 8
x2 - 4
= 2
(b) 2
x - 4 x2 - x - 12
x - x - 12
x3 - 8
=
2
x + 3
# 1x - 22 1x + 2x + 42
1x - 22 1x + 22
1x - 42 1x + 32
=
1x + 32 1x - 22 1x2 + 2x + 42
1x - 22 1x + 22 1x - 42 1x + 32
=
x2 + 2x + 4
1x + 22 1x - 42
Now Work
PROBLEMS
19
x ≠ - 3, - 2, 2, 4
AND
r
27
3 Add and Subtract Rational Expressions
In Words
To add (or subtract) two
rational expressions with the
same denominator, keep the
common denominator and add
(or subtract) the numerators.
The rules for adding and subtracting rational expressions are the same as the rules
for adding and subtracting rational numbers. If the denominators of two rational
expressions to be added (or subtracted) are equal, then add (or subtract) the
numerators and keep the common denominator.
If
a
c
and are two rational expressions, then
b
b
c
a + c
a
+
=
b
b
b
EX AMPLE 4
c
a - c
a
=
b
b
b
if b ≠ 0
(4)
Adding and Subtracting Rational Expressions
with Equal Denominators
Perform the indicated operation and simplify the result. Leave your answer in factored
form.
(a)
Solution
(a)
2x2 - 4
x + 3
+
2x + 5
2x + 5
x ≠ -
5
2
(b)
x
3x + 2
x - 3
x - 3
12x2 - 42 + 1x + 32
2x2 - 4
x + 3
+
=
2x + 5
2x + 5
2x + 5
2
12x - 12 1x + 12
2x + x - 1
=
=
2x + 5
2x + 5
x ≠ 3
SECTION R.7 Rational Expressions
(b)
EXAMPL E 5
Solution
x - 13x + 22
x
3x + 2
x - 3x - 2
=
=
x - 3
x - 3
x - 3
x - 3
21x
+
12
- 2x - 2
=
=
x - 3
x - 3
65
r
Adding Rational Expressions Whose Denominators
Are Additive Inverses of Each Other
Perform the indicated operation and simplify the result. Leave your answer in
factored form.
2x
5
+
x ≠ 3
x - 3
3 - x
Notice that the denominators of the two rational expressions are different. However,
the denominator of the second expression is the additive inverse of the denominator
of the first. That is,
3 - x = - x + 3 = - 1 # 1x - 32 = - 1x - 32
Then
2x
5
2x
5
2x
-5
+
=
+
=
+
x - 3
3 - x
x - 3
- 1x - 32
x - 3
x - 3
c
a c -a
3 - x = - (x - 3)
=
-b
b
2x + 1 - 52
2x - 5
=
=
x - 3
x - 3
Now Work
PROBLEMS
39
AND
r
45
If the denominators of two rational expressions to be added or subtracted are
not equal, we can use the general formulas for adding and subtracting rational
expressions.
a
c
a#d
b#c
ad + bc
+
= # + # =
b
d
b d
b d
bd
#
#
a
c
a d
b c
ad - bc
= # - # =
b
d
b d
b d
bd
EXAMPL E 6
if b ≠ 0, d ≠ 0
(5a)
if b ≠ 0, d ≠ 0
(5b)
Adding and Subtracting Rational Expressions
with Unequal Denominators
Perform the indicated operation and simplify the result. Leave your answer in
factored form.
x
x
x
(a)
x
(a)
Solution
+
+
3
x
x2
1
+
x ≠ - 4, 2
(b) 2
x
4
x - 2
x - 4
3
x
x - 3#x - 2
x + 4# x
+
=
+
4
x - 2
x + 4 x - 2
x + 4 x - 2
æ
(5a)
x ≠ - 2, 0, 2
=
1x - 32 1x - 22 + 1x + 42 1x2
1x + 42 1x - 22
=
2x2 - x + 6
x2 - 5x + 6 + x2 + 4x
=
1x + 42 1x - 22
1x + 42 1x - 22
66
CHAPTER R Review
(b)
x2 1x2 - 1x2 - 42 112
x2
x2 # x
x2 - 4 # 1
1
=
=
x
x2 - 4
x2 - 4 x
x2 - 4 x
1x2 - 42 1x2
æ
(5b)
=
x3 - x2 + 4
1x - 22 1x + 22 1x2
Now Work
PROBLEM
r
49
4 Use the Least Common Multiple Method
If the denominators of two rational expressions to be added (or subtracted) have
common factors, we usually do not use the general rules given by equations (5a)
and (5b). Just as with fractions, we apply the least common multiple (LCM) method.
The LCM method uses the polynomial of least degree that has each denominator
polynomial as a factor.
The LCM Method for Adding or Subtracting Rational
Expressions
The Least Common Multiple (LCM) Method requires four steps:
STEP 1:
STEP 2:
STEP 3:
STEP 4:
Factor completely the polynomial in the denominator of each rational
expression.
The LCM of the denominators is the product of each of these factors
raised to a power equal to the greatest number of times that the factor
occurs in the polynomials.
Write each rational expression using the LCM as the common
denominator.
Add or subtract the rational expressions using equation (4).
We begin with an example that requires only Steps 1 and 2.
EX AMPLE 7
Finding the Least Common Multiple
Find the least common multiple of the following pair of polynomials:
x1x - 12 2 1x + 12
Solution
and 41x - 12 1x + 12 3
STEP 1: The polynomials are already factored completely as
x1x - 12 2 1x + 12
and 41x - 12 1x + 12 3
STEP 2: Start by writing the factors of the left-hand polynomial. (Or you could start
with the one on the right.)
x1x - 12 2 1x + 12
Now look at the right-hand polynomial. Its first factor, 4, does not appear in
our list, so we insert it.
4x1x - 12 2 1x + 12
The next factor, x - 1, is already in our list, so no change is necessary. The
final factor is 1x + 12 3. Since our list has x + 1 to the first power only, we
replace x + 1 in the list by 1x + 12 3. The LCM is
4x1x - 12 2 1x + 12 3
SECTION R.7 Rational Expressions
67
Notice that the LCM is, in fact, the polynomial of least degree that contains
x1x - 12 2 1x + 12 and 41x - 12 1x + 12 3 as factors.
r
Now Work
PROBLEM
55
Using the Least Common Multiple to Add Rational Expressions
EXAMPL E 8
Perform the indicated operation and simplify the result. Leave your answer in factored
form.
x
2x - 3
+ 2
x2 + 3x + 2
x - 1
Solution
x ≠ - 2, - 1, 1
STEP 1: Factor completely the polynomials in the denominators.
x2 + 3x + 2 = 1x + 22 1x + 12
x2 - 1 = 1x - 12 1x + 12
STEP 2: The LCM is 1x + 22 1x + 12 1x - 12 . Do you see why?
STEP 3: Write each rational expression using the LCM as the denominator.
x1x - 12
x
x
x
#x - 1 =
=
=
1x + 22 1x + 12
1x + 22 1x + 12 x - 1
1x + 22 1x + 12 1x - 12
x + 3x + 2
2
æ Multiply numerator and
denominator by x - 1 to get
the LCM in the denominator.
2x - 3
2x - 3
2x - 3
# x + 2 = 12x - 32 1x + 22
=
=
2
1x - 12 1x + 12
1x - 12 1x + 12 x + 2
1x - 12 1x + 12 1x + 22
x - 1
æ Multiply numerator and
denominator by x + 2 to get
the LCM in the denominator.
STEP 4: Now add by using equation (4).
12x - 32 1x + 22
x1x - 12
x
2x - 3
+
+ 2
=
1x + 22 1x + 12 1x - 12
1x + 22 1x + 12 1x - 12
x2 + 3x + 2
x - 1
EXAMPL E 9
=
1x2 - x2 + 12x2 + x - 62
1x + 22 1x + 12 1x - 12
=
31x2 - 22
3x2 - 6
=
1x + 22 1x + 12 1x - 12
1x + 22 1x + 12 1x - 12
r
Using the Least Common Multiple to Subtract Rational Expressions
Perform the indicated operation and simplify the result. Leave your answer in factored
form.
3
x + 4
- 2
x2 + x
x + 2x + 1
Solution
x ≠ - 1, 0
STEP 1: Factor completely the polynomials in the denominators.
x2 + x = x1x + 12
x2 + 2x + 1 = 1x + 12 2
STEP 2: The LCM is x1x + 12 2.
68
CHAPTER R Review
STEP 3: Write each rational expression using the LCM as the denominator.
3
3
3
# x + 1 = 31x + 122
=
=
x1x + 12
x1x + 12 x + 1
x + x
x1x + 12
2
x1x + 42
x + 4
x + 4
x + 4 #x
=
=
=
2
2 x
x + 2x + 1
1x + 12
1x + 12
x1x + 12 2
2
STEP 4: Subtract, using equation (4).
31x + 12
x1x + 42
3
x + 4
- 2
=
2
x + x
x + 2x + 1
x1x + 12
x1x + 12 2
2
=
Now Work
PROBLEM
31x + 12 - x1x + 42
x1x + 12 2
=
3x + 3 - x2 - 4x
x1x + 12 2
=
- x2 - x + 3
x1x + 12 2
r
65
5 Simplify Complex Rational Expressions
When sums and/or differences of rational expressions appear as the numerator
and/or denominator of a quotient, the quotient is called a complex rational expression.*
For example,
1
x
1
1 x
1 +
and
x2
- 3
x2 - 4
x - 3
- 1
x + 2
are complex rational expressions. To simplify a complex rational expression means
to write it as a rational expression reduced to lowest terms. This can be accomplished
in either of two ways.
Simplifying a Complex Rational Expression
OPTION 1: Treat the numerator and denominator of the complex rational
expression separately, performing whatever operations are indicated
and simplifying the results. Follow this by simplifying the resulting
rational expression.
OPTION 2: Find the LCM of the denominators of all rational expressions that
appear in the complex rational expression. Multiply the numerator
and denominator of the complex rational expression by the LCM
and simplify the result.
We use both options in the next example. By carefully studying each option, you
can discover situations in which one may be easier to use than the other.
EX A MPL E 1 0
Simplifying a Complex Rational Expression
1
3
+
x
2
Simplify:
x + 3
4
x ≠ - 3, 0
* Some texts use the term complex fraction.
SECTION R.7 Rational Expressions
Solution
69
Option 1: First, we perform the indicated operation in the numerator, and then we
divide.
1
3
1#x + 2#3
x + 6
+
x
2
2#x
2x
x + 6# 4
=
=
=
x + 3
x + 3
x + 3
2x
x + 3
æ
4 æ
4
4
Rule for adding quotients
=
Rule for dividing quotients
1x + 62 # 4
2 # 2 # 1x + 62
21x + 62
=
=
2 # x # 1x + 32
2 # x # 1x + 32
x1x + 32
æ
Rule for multiplying quotients
Option 2: The rational expressions that appear in the complex rational expression
are
1
,
2
3
,
x
x + 3
4
The LCM of their denominators is 4x. We multiply the numerator and
denominator of the complex rational expression by 4x and then simplify.
1
3
3
1
3
1
4x # a + b
+
4x # + 4x #
x
x
x
2
2
2
=
=
#
4x 1x + 32
x + 3
x + 3
4x # a
b
4
4
4
æ Multiply the
æ Use the Distributive Property
numerator and
in the numerator.
denominator by 4x.
1
3
+ 4x #
21x + 62
x
2x + 12
2
=
=
#
4 x 1x + 32
x1x + 32
x1x + 32
æ
æ
4
2 # 2x #
=
Simplify
EX AM PL E 11
r
Simplifying a Complex Rational Expression
x2
+ 2
x - 4
Simplify:
2x - 2
- 1
x
Solution
Factor
x ≠ 0, 2, 4
We will use Option 1.
21x - 42
x2
x2 + 2x - 8
x2
+ 2
+
x - 4
x - 4
x - 4
x - 4
=
=
2x - 2
2x - 2
x
2x - 2 - x
- 1
x
x
x
x
1x + 42 1x - 22
1x + 42 1x - 22
x - 4
=
=
x - 2
x - 4
x
=
Now Work
1x + 42 # x
x - 4
PROBLEM
75
#
x
x - 2
r
70
CHAPTER R Review
Application
Solving an Application in Electricity
EX A MPL E 12
An electrical circuit contains two resistors connected in parallel, as shown in
Figure 28. If these two resistors provide resistance of R1 and R2 ohms, respectively,
their combined resistance R is given by the formula
R1
R =
R2
Figure 28
1
1
1
+
R1
R2
Express R as a rational expression; that is, simplify the right-hand side of this formula.
Evaluate the rational expression if R1 = 6 ohms and R2 = 10 ohms.
Solution
1
We will use Option 2. If we consider 1 as the fraction , the rational expressions in
1
the complex rational expression are
1
1
1
,
,
1 R1 R2
The LCM of the denominators is R1 R2 . We multiply the numerator and denominator
of the complex rational expression by R1 R2 and simplify.
1
1
1
+
R1
R2
=
1 # R1 R2
a
1
1 #
+
b R1 R2
R1
R2
=
R1 R2
1 #
1 #
RR +
RR
R1 1 2
R2 1 2
=
R1 R2
R2 + R1
So,
R =
R1 R2
R2 + R1
If R1 = 6 and R2 = 10, then
R =
6 # 10
60
15
=
=
10 + 6
16
4
ohms
r
R.7 Assess Your Understanding
Concepts and Vocabulary
1. When the numerator and denominator of a rational
expression contain no common factors (except 1 and - 1), the
rational expression is in
.
2. LCM is an abbreviation for
.
3. Choose the statement that is not true. Assume b ≠ 0, c ≠ 0,
and d ≠ 0 as necessary.
a
a
c
a + c
ac
(a)
=
(b)
+
=
bc
b
b
b
b
a
a
b
ac
c
ad - bc
(c)
=
=
(d)
b
d
bd
c
bd
d
4. Choose the rational expression that simplifies to - 1.
a - b
a - b
(a)
(b)
b - a
a - b
a + b
b - a
(c)
(d)
a - b
b + a
2x3 - 4x
is reduced
5. True or False The rational expression
x - 2
to lowest terms.
6. True or False The LCM of 2x3 + 6x2 and 6x4 + 4x3 is
4x3(x + 1).
Skill Building
In Problems 7–18, reduce each rational expression to lowest terms.
7.
3x + 9
x2 - 9
8.
4x2 + 8x
12x + 24
9.
11.
24x2
12x2 - 6x
12.
x2 + 4x + 4
x2 - 4
13.
15.
x2 + 4x - 5
x2 - 2x + 1
16.
x - x2
x2 + x - 2
17.
x2 - 2x
3x - 6
y2 - 25
2
2y - 8y - 10
x2 + 5x - 14
2 - x
10.
14.
18.
15x2 + 24x
3x2
3y2 - y - 2
3y2 + 5y + 2
2x2 + 5x - 3
1 - 2x
SECTION R.7 Rational Expressions
71
In Problems 19–36, perform the indicated operation and simplify the result. Leave your answer in factored form.
19.
3x + 6 # x
5x2
x2 - 4
23.
12
4x - 8 #
- 3x 12 - 6x
20.
3 #
x2
2x 6x + 10
24.
21.
4x2 # x3 - 64
2x
x2 - 16
2
6x - 27 #
5x
4x - 18
25.
6x
x - 4
27.
3x - 9
2x + 4
x2 + x - 6 #
x2 - 25
26. 2
x + 4x - 5 x2 + 2x - 15
2
28.
30.
x2 + 7x + 12
x2 - 7x + 12
33. 2
x + x - 12
x2 - x - 12
x2 + 7x + 6
x2 + x - 6
34. 2
x + 5x - 6
x2 + 5x + 6
12 # x3 + 1
x2 + x 4x - 2
x2 - 3x - 10 # x2 + 4x - 21
x2 + 2x - 35 x2 + 9x + 14
12x
5x + 20
4x2
x - 16
2
x - 2
4x
8x
x2 - 1
29.
10x
x + 1
22.
x2 - 4x + 4
12x
4 - x
4 + x
31.
4x
2
x - 16
32.
2x2 - x - 28
3x2 - x - 2
35.
4x2 + 16x + 7
3x2 + 11x + 6
9x2 + 3x - 2
12x2 + 5x - 2
36.
9x2 - 6x + 1
8x2 - 10x - 3
3 + x
3 - x
x2 - 9
9x3
In Problems 37–54, perform the indicated operation and simplify the result. Leave your answer in factored form.
37.
5
x
+
2
2
38.
6
3
x
x
39.
x2
4
2x - 3
2x - 3
40.
9
3x2
2x - 1
2x - 1
41.
x + 1
2x - 3
+
x - 3
x - 3
42.
2x - 5
x + 4
+
3x + 2
3x + 2
43.
3x + 5
2x - 4
2x - 1
2x - 1
44.
5x - 4
x + 1
3x + 4
3x + 4
45.
x
4
+
x - 2
2 - x
46.
6
x
x - 1
1 - x
47.
4
2
x - 1
x + 2
48.
5
2
x + 5
x - 5
49.
2x - 3
x
+
x + 1
x - 1
50.
3x
2x
+
x - 4
x + 3
51.
x - 3
x + 4
x + 2
x - 2
52.
2x - 3
2x + 1
x - 1
x + 1
53.
1
x
+
x
x2 - 4
54.
x - 1
x
+ 2
x + 1
x3
In Problems 55–62, find the LCM of the given polynomials.
55. x2 - 4, x2 - x - 2
56. x2 - x - 12, x2 - 8x + 16
57. x3 - x, x2 - x
58. 3x2 - 27, 2x2 - x - 15
59. 4x3 - 4x2 + x, 2x3 - x2, x3
60. x - 3, x2 + 3x, x3 - 9x
61. x3 - x, x3 - 2x2 + x, x3 - 1
62. x2 + 4x + 4, x3 + 2x2,
1x + 22 3
In Problems 63–74, perform the indicated operations and simplify the result. Leave your answer in factored form.
63.
x
x
- 2
x2 - 7x + 6
x - 2x - 24
64.
66.
x - 4
3x
- 2
x - 1
x - 2x + 1
67.
1x - 12 1x + 12
69.
2x + 3
x + 4
- 2
x2 - x - 2
x + 2x - 8
70.
72.
x + 1
x
2
+ - 3
x
1x - 12 2
x - x2
73.
x
x + 1
- 2
x - 3
x + 5x - 24
65.
2
4x
- 2
x2 - 4
x + x - 6
68.
6
2
1x + 22 2 1x - 12
1x + 22 1x - 12 2
2x - 3
x - 2
x2 + 8x + 7
1x + 12 2
71.
3
1
2
+ 3
- 2
x
x + x
x - x2
1
1
1
a
- b
h x + h
x
74.
1
1
1
J
- 2R
h 1x + h2 2
x
3
2
+
2
1x - 12 1x + 12 2
72
CHAPTER R Review
In Problems 75–86, perform the indicated operations and simplify the result. Leave your answer in factored form.
x
x + 1
78.
x - 1
2 x
1
x
75.
1
1 x
1
x2
76.
1
3 - 2
x
x + 1
x
77.
x - 1
3 +
x + 1
x - 3
x + 4
x - 2
x + 1
79.
x + 1
x - 2
x
x + 1
x - 2
80.
x + 3
x - 2
x - 1
+
x + 2
x + 1
81.
x
2x - 3
x + 1
x
1 +
83. 1 -
4 +
1
1 -
1
84. 1 -
1
x
1 -
2 -
1 -
85.
1
1 - x
82.
21x - 12 - 1 + 3
31x - 12
-1
86.
+ 2
2x + 5
x
x
x - 3
1x + 12 2
x2
x - 3
x + 3
41x + 22 - 1 - 3
31x + 22 - 1 - 1
In Problems 87–94, expressions that occur in calculus are given. Reduce each expression to lowest terms.
87.
90.
93.
12x + 32 # 3 - 13x - 52 # 2
13x - 52 2
88.
x # 2x - 1x2 - 42 # 1
1x2 - 42
91.
2
14x + 12 # 5 - 15x - 22 # 4
15x - 22 2
13x + 12 # 2x - x2 # 3
13x + 12 2
1x2 + 12 # 3 - 13x + 42 # 2x
1x2 + 12
89.
92.
1x2 + 12
2
12x - 52 # 3x2 - x3 # 2
12x - 52 2
1x2 + 92 # 2 - 12x - 52 # 2x
94.
2
x # 2x - 1x2 + 12 # 1
1x2 + 92
2
Applications and Extensions
96. Electrical Circuits An electrical circuit contains three
resistors connected in parallel. If these three resistors provide
resistance of R1 , R2 , and R3 ohms, respectively, their
combined resistance R is given by the formula
95. The Lensmaker’s Equation The focal length f of a lens with
index of refraction n is
1
1
1
= 1n - 12 J
+
R
f
R1
R2
1
1
1
1
=
+
+
R
R1
R2
R3
where R1 and R2 are the radii of curvature of the front and
back surfaces of the lens. Express f as a rational expression.
Evaluate the rational expression for n = 1.5, R1 = 0.1 meter,
and R2 = 0.2 meter.
Express R as a rational expression. Evaluate R for
R1 = 5 ohms, R2 = 4 ohms, and R3 = 10 ohms.
Explaining Concepts: Discussion and Writing
97. The following expressions are called continued fractions:
1 +
1
, 1 +
x
1
1 +
1
x
1
, 1 +
1 +
1
1 +
1
, 1 +
1
x
,...
1
1 +
1 +
1
1 +
1
x
Each simplifies to an expression of the form
ax + b
bx + c
Trace the successive values of a, b, and c as you “continue” the fraction. Can you discover the patterns that these values follow? Go
to the library and research Fibonacci numbers. Write a report on your findings.
98. Explain to a fellow student when you would use the LCM
method to add two rational expressions. Give two examples
of adding two rational expressions: one in which you use the
LCM and the other in which you do not.
99. Which of the two options given in the text for simplifying
complex rational expressions do you prefer? Write a brief
paragraph stating the reasons for your choice.
SECTION R.8 nth Roots; Rational Exponents
73
R.8 nth Roots; Rational Exponents
PREPARING FOR THIS SECTION Before getting started, review the following:
r Exponents, Square Roots (Section R.2, pp. 21–24)
Now Work the ‘Are You Prepared?’ problems on page 78.
OBJECTIVES 1 Work with nth Roots (p. 73)
2 Simplify Radicals (p. 74)
3 Rationalize Denominators (p. 75)
4 Simplify Expressions with Rational Exponents (p. 76)
1 Work with nth Roots
DEFINITION
The principal nth root of a real number a, n Ú 2 an integer, symbolized by
n
2a, is defined as follows:
n
2a = b means a = bn
where a Ú 0 and b Ú 0 if n is even and a, b are any real numbers if n is odd.
n
In Words
n
The symbol 2a means “give me
the number that, when raised to
the power n, equals a.”
EXAMPL E 1
Notice that if a is negative and n is even, then 2a is not defined. When it
is defined, the principal nth root of a number is unique.
n
The symbol 2a for the principal nth root of a is called a radical; the integer n
is called the index, and a is called the radicand. If the index of a radical is 2, we call
2
2
a the square root of a and omit the index 2 by simply writing 1a. If the index is
3
3, we call 2
a the cube root of a.
Simplifying Principal nth Roots
3
3 3
(a) 2
8 = 2
2 = 2
(c)
3
3
(b) 2
- 64 = 2
1 - 42 3 = - 4
6
(d) 2
1 - 22 6 = 0 - 2 0 = 2
1
1 4
1
= 4 a b =
A 16
2
B 2
4
r
These are examples of perfect roots, since each simplifies to a rational number.
Notice the absolute value in Example 1(d). If n is even, then the principal nth root
must be nonnegative.
In general, if n Ú 2 is an integer and a is a real number, we have
n
2a n = a
2a n = 0 a 0
n
Now Work
PROBLEM
if n Ú 3 is odd
(1a)
if n Ú 2 is even
(1b)
11
Radicals provide a way of representing many irrational real numbers. For
example, it can be shown that there is no rational number whose square is 2. Using
radicals, we can say that 12 is the positive number whose square is 2.
74
CHAPTER R Review
EX AMPLE 2
Using a Calculator to Approximate Roots
5
Use a calculator to approximate 2
16.
Solution Figure 29 shows the result using a TI-84 Plus C graphing calculator.
Now Work
r
113
PROBLEM
2 Simplify Radicals
Let n Ú 2 and m Ú 2 denote integers, and let a and b represent real numbers.
Assuming that all radicals are defined, we have the following properties:
Figure 29
Properties of Radicals
n
n
n
2ab = 2a 2b
(2a)
n
2a
n a
= n
Ab
2b
n
b ≠ 0
n
2am = ( 2a2 m
(2b)
(2c)
When used in reference to radicals, the direction to “simplify” will mean to
remove from the radicals any perfect roots that occur as factors.
EX AMPLE 3
Simplifying Radicals
(a) 232 = 216 # 2 = 216 # 22 = 422
c
c
(2a)
Factor out 16,
a perfect square.
3
3
3
3
3 3# 3
3
(b) 2
16 = 2
8#2 = 2
8# 2
2 = 2
2 22 = 22
2
c
c
Factor out 8,
a perfect cube.
(2a)
3
3
3
(c) 2
- 16x4 = 2
- 8 # 2 # x3 # x = 2
1 - 8x3 2 12x2
c
c
Group perfect
Factor perfect
cubes inside radical. cubes.
3
3
3
3
= 2
1 - 2x2 3 # 2x = 2
1 - 2x2 3 # 2
2x = - 2x2
2x
c
(2a)
(d)
16x5
24 x4 x
2x 4 #
2x 4 # 4
2x 4
4
4
= 4
=
a
b
x
=
a
b 2x = ` ` 2
x
4
3
B 81
B 3
B 3
B 3
4
Now Work
PROBLEMS
15
AND
21
r
Two or more radicals can be combined, provided that they have the same index
and the same radicand. Such radicals are called like radicals.
EX AMPLE 4
Combining Like Radicals
(a) - 8212 + 23 = - 824 # 3 + 23
= - 8 # 24 23 + 23
= - 1623 + 23 = - 1523
SECTION R.8 nth Roots; Rational Exponents
75
3
3
3
3 3 3
3
3 3
(b) 2
8x4 + 2
- x + 42
27x = 2
2xx + 2
- 1 # x + 42
3 x
3
3
3
3
3 3# 3
= 2
12x2 3 # 2
x + 2
-1 # 2
x + 42
3 2x
3
3
3
= 2x2
x - 1# 2
x + 122
x
3
= 12x + 112 2
x
Now Work
PROBLEM
r
39
3 Rationalize Denominators
When radicals occur in quotients, it is customary to rewrite the quotient so that the
new denominator contains no radicals. This process is referred to as rationalizing
the denominator.
The idea is to multiply by an appropriate expression so that the new denominator
contains no radicals. For example:
If a Denominator
Contains the Factor
Multiply by
To Obtain a Denominator
Free of Radicals
23
23
23 + 1
23 - 1
22 - 3
22 + 3
25 - 23
25 + 23
1 252 - 1 232 2 = 5 - 3 = 2
3
2
4
3
2
2
3
3
3
2
4# 2
2 = 2
8 = 2
1 232 = 3
2
1 232 - 12 = 3 - 1 = 2
2
1 222 - 32 = 2 - 9 = - 7
2
2
In rationalizing the denominator of a quotient, be sure to multiply both the numerator
and the denominator by the expression.
EXAMPL E 5
Solution
Rationalizing Denominators
Rationalize the denominator of each expression:
1
5
22
(a)
(b)
(c)
23
422
23 - 322
(a) The denominator contains the factor 23, so we multiply the numerator and
denominator by 23 to obtain
1
23
=
1
# 23
23 23
=
23
1 232
2
=
23
3
(b) The denominator contains the factor 12, so we multiply the numerator and
denominator by 12 to obtain
5
422
=
5
# 22
422 22
=
522
41 222
2
=
522
522
=
4#2
8
(c) The denominator contains the factor 23 - 322, so we multiply the numerator
and denominator by 23 + 322 to obtain
22
23 - 322
=
# 23 + 322
22
23 - 322 23 + 322
=
221 23 + 3222
1 232 - 13222 2
2
2
=
Now Work
22 23 + 31 222
26 + 6
6 + 26
=
= 3 - 18
- 15
15
PROBLEM
53
r
76
CHAPTER R Review
4 Simplify Expressions with Rational Exponents
Radicals are used to define rational exponents.
DEFINITION
If a is a real number and n Ú 2 is an integer, then
n
a1>n = 2a
(3)
n
provided that 2a exists.
n
Note that if n is even and a 6 0, then 2a and a1>n do not exist.
EX AMPLE 6
Writing Expressions Containing Fractional Exponents as Radicals
(a) 41>2 = 24 = 2
(b) 81>2 = 28 = 222
3
(c) 1 - 272 1>3 = 2
- 27 = - 3
DEFINITION
r
3
3
(d) 161>3 = 2
16 = 22
2
If a is a real number and m and n are integers containing no common factors,
with n Ú 2, then
am>n = 2am = 1 2a2 m
n
n
(4)
n
provided that 2a exists.
We have two comments about equation (4):
m
must be in lowest terms, and n Ú 2 must be positive.
n
n
n
2. In simplifying the rational expression am>n, either 2am or 1 2a2 m may
be used, the choice depending on which is easier to simplify. Generally, taking the
n
root first, as in 1 2a2 m, is easier.
1. The exponent
EX AMPLE 7
Using Equation (4)
(a) 43>2 = 1 242 = 23 = 8
3
(b) 1 - 82 4>3 = 1 2
- 82 = 1 - 22 4 = 16
3
4
5
(c) 1322 -2>5 = 1 2
322 -2 = 2-2 =
Now Work
PROBLEM
1
3
(d) 256>4 = 253>2 = 1 2252 = 53 = 125
4
r
63
It can be shown that the Laws of Exponents hold for rational exponents. The
next example illustrates using the Laws of Exponents to simplify.
EX AMPLE 8
Simplifying Expressions Containing Rational Exponents
Simplify each expression. Express your answer so that only positive exponents
occur. Assume that the variables are positive.
(a) 1x2>3 y2 1x -2 y2
1>2
(b) ¢
2x1>3
y
2>3
≤
-3
(c) ¢
9x2 y1>3
x
1>3
y
≤
1>2
SECTION R.8 nth Roots; Rational Exponents
Solution
(a) 1x2>3 y2 1x -2 y2
1>2
= 1x2>3 y2 3 1x -2 2
= x
2>3
= 1x
1>2 1>2
y
77
4
-1 1>2
yx y
2>3
# x-1 2 1y # y1>2 2
= x -1>3 y3>2
=
(b) ¢
(c) ¢
2x1>3
y2>3
≤
-3
9x2 y1>3
x1>3 y
≤
= ¢
1>2
y2>3
2x
= ¢
Now Work
y3>2
x1>3
3
≤ =
1>3
9x2 - 11>32
y1 - 11>32
1y2>3 2
3
12x1>3 2
≤
1>2
PROBLEM
3
= ¢
=
y2
23 1x1>3 2
9x5>3
y2>3
≤
1>2
3
=
=
y2
8x
91>2 1x5>3 2
1y2>3 2
1>2
1>2
=
3x5>6
y1>3
r
83
The next two examples illustrate some algebra that you will need to know for
certain calculus problems.
EXAMPL E 9
Writing an Expression as a Single Quotient
Write the following expression as a single quotient in which only positive exponents
appear.
1x2 + 12
Solution
1x2 + 12
1>2
+ x#
1>2
+ x#
1 2
1x + 12 -1>2 # 2x
2
1 2
x2
1>2
- 1>2 #
1x + 12
2 x = 1x2 + 12
+
1>2
2
1x2 + 12
=
=
=
Now Work
EX AM PL E 1 0
1>2
1x2 + 12
1x2 + 12
1>2
+ x2
1>2
1x2 + 12 + x2
1x2 + 12
1>2
2x2 + 1
1x2 + 12
1>2
r
89
Factoring an Expression Containing Rational Exponents
Factor and simplify:
Solution
PROBLEM
1x2 + 12
4 1>3
x 12x + 12 + 2x4>3
3
Begin by writing 2x4>3 as a fraction with 3 as the denominator.
4x1>3(2x + 1)
4x1>3(2x + 1) + 6x4>3
4 1>3
6x4>3
x 12x + 12 + 2x4>3 =
+
=
3
3
3 c
3
Add the two fractions
2x1>3[2(2x + 1) + 3x]
2x1>3(7x + 2)
=
3
3
c
c
=
2 and x 1>3 are common factors
Now Work
PROBLEM
101
Simplify
r
78
CHAPTER R Review
Historical Note
T
he radical sign, 2 , was first used in print by Christoff Rudolff in
1525. It is thought to be the manuscript form of the letter r (for
the Latin word radix = root), although this has not been quite
conclusively confirmed. It took a long time for 2 to become the
standard symbol for a square root and much longer to standardize
3
4
5
2
, 2, 2 , and so on. The indexes of the root were placed in every
conceivable position, with
3
2 8,
3 8, and
2○
3
4
all being variants for 2
8 . The notation 2 216 was popular for 2
16 .
By the 1700s, the index had settled where we now put it.
The bar on top of the present radical symbol, as follows,
2a2 + 2ab + b2
is the last survivor of the vinculum, a bar placed atop an expression to
indicate what we would now indicate with parentheses. For example,
28
ab + c = a1b + c2
3
R.8 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages in red.
1. 1 - 32 2 =
; - 32 =
2. 216 =
(pp. 21–24)
; 21 - 42 2 =
(pp. 21–24)
Concepts and Vocabulary
n
3. In the symbol 2a, the integer n is called the
3
4. We call 2a the
.
of a.
5. Let n Ú 2 and m Ú 2 be integers, and let a and b be real
numbers. Which of the following is not a property of
radicals? Assume all radicals are defined.
n
(a)
2a
a
= n
Ab
2b
n
n
n
n
n
7. Which of the following phrases best defines like radicals?
(a) Radical expressions that have the same index
(b) Radical expressions that have the same radicand
(c) Radical expressions that have the same index and the
same radicand
(d) Radical expressions that have the same variable
12
,
1 - 13
multiply both the numerator and the denominator by which
n
(b) 2a + b = 2a + 2b
8. To rationalize the denominator of the expression
(d) 2am = 1 2a2 m
n
n
(c) 2ab = 2a2b
n
6. If a is a real number and n Ú 2 is an integer, then which of
n
the following expressions is equivalent to 2a, provided that
it exists?
1
(a) a-n
(b) an
(c)
(d) a1>n
an
of the following?
(b) 22
(a) 23
(c) 1 + 23
(d) 1 - 23
5
9. True or False 2 - 32 = - 2
4
10. True or False 2 1 - 32 4 = - 3
Skill Building
In Problems 11–48, simplify each expression. Assume that all variables are positive when they appear.
3
11. 227
4
12. 2
16
3
13. 2
-8
15. 28
3
16. 2
54
17. 2 - 8x4
4 12 8
19. 2
x y
5 10 5
20. 2
x y
21.
23. 236x
24. 29x5
4
25. 2162x9 y12
3
26. 2
- 40x14 y10
27. 23x2 212x
28. 25x 220x3
3
29. 1 25 2
92
3
30. 1 2
3 2102
31. 13262 12222
32. 15282 1 - 3232
33. 322 + 422
34. 625 - 425
35. - 218 + 228
36. 2212 - 3227
37. 1 23 + 32 1 23 - 12
38. 1 25 - 22 1 25 + 32
3
3
39. 52
2 - 22
54
3
3
40. 92
24 - 2
81
41. 1 2x - 12 2
42. 1 2x + 252 2
3
3
43. 216x4 - 22x
4
4
44. 2
32x + 2
2x5
45. 28x3 - 3250x
3
3
3
47. 216x4 y - 3x22xy + 52 - 2xy4
3
14. 2
-1
3
4
4
18. 2
48x5
x9 y 7
B xy
22.
3
2
3
3xy2
B 81x4 y2
4
46. 3x29y + 4225y
3
48. 8xy - 225x2 y2 + 28x3 y3
SECTION R.8
nth Roots; Rational Exponents
79
In Problems 49–62, rationalize the denominator of each expression. Assume that all variables are positive when they appear.
49.
53.
57.
61.
1
2
50.
22
23
22
54.
5 - 22
5
55.
27 + 2
- 23
52.
25
2 - 25
56.
2 + 325
-3
58.
22 - 1
51.
23
59.
25 + 4
2x + h - 2x
62.
2x + h + 2x
In Problems 63–78, simplify each expression.
- 23
28
23 - 1
223 + 3
5
60.
3
22
-2
3
2
9
2x + h + 2x - h
2x + h - 2x - h
63. 82>3
64. 43>2
65. 1 - 272 1>3
66. 163>4
67. 163>2
68. 253>2
69. 9-3>2
70. 16 -3>2
9 3>2
71. a b
8
72. a
8 -3>2
73. a b
9
74. a
75. 1 - 10002 -1>3
76. - 25-1>2
27 2>3
b
8
77. a -
64 -2>3
b
125
8 -2>3
b
27
78. - 81-3>4
In Problems 79–86, simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are
positive.
79. x3>4 x1>3 x -1>2
83.
1x2 y2
1>3
81. 1x3 y6 2
80. x2>3 x1>2 x -1>4
1xy2 2
1xy2 1>4 1x2 y2 2 1>2
2>3
1x2y2 3>4
84.
x2>3 y2>3
85.
82. 1x4y8 2
1>3
116x2 y -1>3 2
3>4
1xy2 2 1>4
86.
3>4
14x -1 y1>3 2
3>2
1xy2 3>2
Applications and Extensions
In Problems 87–100, expressions that occur in calculus are given. Write each expression as a single quotient in which only positive
exponents and/or radicals appear.
87.
x
11 + x2
+ 211 + x2 1>2
1>2
89. 2x1x2 + 12 1>2 + x2 #
91. 24x + 3 #
1
22x - 5
21 + x - x #
95.
97.
1x2 - 12
1>2
- 1x2 - 12
1
524x + 3
x 7 5
92.
x 7 -1
x 7 -4
94.
96.
1>2
x 6 - 1 or x 7 1
98.
2
+ x1>2
x 7 0
3
2
8x + 1
32 1x - 22
3
2
+
1
1x + 12 -2>3
3
x 7 0
100.
x ≠ -1
3
2
x - 2
242 18x + 12
3
2
x ≠ 2, x ≠ -
1
8
2x
22x2 + 1
x + 1
2
19 - x2 2 1>2 + x2 19 - x2 2 -1>2
9 - x2
-3 6 x 6 3
1x2 + 42 1>2 - x2 1x2 + 42 -1>2
x2 + 4
2x11 - x2 2 1>3 +
- 2x2x
11 + x2 2
2x1>2
2x2 + 1 - x #
x2
22x
1 + x
90. 1x + 12 1>3 + x #
1
x + 4
1 + x2
99.
+ 2x - 5 #
1x + 42 1>2 - 2x1x + 42 -1>2
x2
88.
1 2
1x + 12 -1>2 # 2x
2
221 + x
1 + x
93.
x 7 -1
2 3
x 11 - x2 2 -2>3
3
11 - x2 2 2>3
x ≠ - 1, x ≠ 1
80
CHAPTER R Review
In Problems 101–110, expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive
exponents occur.
3
4
4>3
1>3
3>2
101. 1x + 12
+ x # 1x + 12 1>2
x Ú -1
102. 1x2 + 42
+ x # 1x2 + 42 # 2x
2
3
103. 6x1>2 1x2 + x2 - 8x3>2 - 8x1>2
105. 31x2 + 42
4>3
+ x # 41x2 + 42
104. 6x1>2 12x + 32 + x3>2 # 8
x Ú 0
1>3
106. 2x13x + 42 4>3 + x2 # 413x + 42 1>3
# 2x
107. 413x + 52 1>3 12x + 32 3>2 + 313x + 52 4>3 12x + 32 1>2 x Ú 109. 3x -1>2 +
3 1>2
x
2
x Ú 0
3
3
108. 616x + 12 1>3 14x - 32 3>2 + 616x + 12 4>3 14x - 32 1>2 x Ú
2
4
110. 8x1>3 - 4x -2>3
x 7 0
x ≠ 0
In Problems 111–118, use a calculator to approximate each radical. Round your answer to two decimal places.
111. 22
115.
2 + 23
3 - 25
3
113. 2
4
112. 27
116.
25 - 2
22 + 4
117.
3
114. 2
-5
3
32
5 - 22
118.
23
3
223 - 2
4
22
119. Calculating the Amount of Gasoline in a Tank A Shell station stores its gasoline in underground tanks that are right circular
cylinders lying on their sides. See the illustration. The volume V of gasoline in the tank (in gallons) is given by the formula
V = 40h2
96
- 0.608
Ah
where h is the height of the gasoline (in inches) as measured on a depth stick.
(a) If h = 12 inches, how many gallons of gasoline are in the tank?
(b) If h = 1 inch, how many gallons of gasoline are in the tank?
120. Inclined Planes The final velocity v of an object in feet per second (ft>sec) after it slides down a frictionless inclined plane of
height h feet is
v = 264h + v20
where v0 is the initial velocity (in ft/sec) of the object.
(a) What is the final velocity v of an object that slides down a frictionless inclined
plane of height 4 feet? Assume that the initial velocity is 0.
(b) What is the final velocity v of an object that slides down a frictionless inclined
plane of height 16 feet? Assume that the initial velocity is 0.
(c) What is the final velocity v of an object that slides down a frictionless inclined
plane of height 2 feet with an initial velocity of 4 ft/sec?
v0
h
v
Problems 121 and 122 require the following information.
Period of a Pendulum The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula
T = 2p
l
A 32
In Problems 121 and 122, express your answer both as a square root and as a decimal.
121. Find the period T of a pendulum whose length is 64 feet.
122. Find the period T of a pendulum whose length is 16 feet.
Explaining Concepts: Discussion and Writing
123. Give an example to show that 2a2 is not equal to a. Use it to explain why 2a2 = 0 a 0 .
‘Are You Prepared?’ Answers
1. 9; - 9
2. 4; 4
Equations and
Inequalities
Financing a Purchase
1
Whenever we make a major purchase, such as an automobile
auto
au
t mobile or a house,
money from a lending
we often need to finance the purchase by borrowing money
institution, such as a bank. Have you ever wondered how
how the bank
terest
determines the monthly payment? How much total interest
will be paid over the course of the loan? What roles do th
the
he
rate of interest and the length of the loan play?
—See the Internet-based Chapter Project I—
—
A Look Ahead
Chapter 1, Equations and Inequalities, reviews many topics covered in Intermediate
Algebra. If your instructor decides to exclude complex numbers from the course,
don’t be alarmed. The book has been designed so that the topic of complex
numbers can be included or excluded without any confusion later on.
Outline
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Linear Equations
Quadratic Equations
Complex Numbers; Quadratic
Equations in the Complex Number
System
Radical Equations; Equations
Quadratic in Form; Factorable
Equations
Solving Inequalities
Equations and Inequalities Involving
Absolute Value
Problem Solving: Interest, Mixture,
Uniform Motion, Constant Rate
Job Applications
Chapter Review
Chapter Test
Chapter Projects
81
81
82
CHAPTER 1 Equations and Inequalities
1.1 Linear Equations
PREPARING FOR THIS SECTION Before getting started, review the following:
r Properties of Real Numbers (Section R.1, pp. 9–13)
r Domain of a Variable (Section R.2, p. 21)
Now Work the ‘Are You Prepared?’ problems on page 90.
OBJECTIVES 1 Solve a Linear Equation (p. 84)
2 Solve Equations That Lead to Linear Equations (p. 86)
3 Solve Problems That Can Be Modeled by Linear Equations (p. 87)
An equation in one variable is a statement in which two expressions, at least one
containing the variable, are set equal. The expressions are called the sides of the
equation. Since an equation is a statement, it may be true or false, depending on
the value of the variable. Unless otherwise restricted, the admissible values of the
variable are those in the domain of the variable. These admissible values of the
variable, if any, that result in a true statement are called solutions, or roots, of
the equation. To solve an equation means to find all the solutions of the equation.
For example, the following are all equations in one variable, x:
x + 5 = 9
x2 + 5x = 2x - 2
x2 - 4
= 0
x + 1
2x2 + 9 = 5
The first of these statements, x + 5 = 9, is true when x = 4 and false for any
other choice of x. That is, 4 is a solution of the equation x + 5 = 9. We also say
that 4 satisfies the equation x + 5 = 9, because, when 4 is substituted for x, a true
statement results.
Sometimes an equation will have more than one solution. For example, the
equation
x2 - 4
= 0
x + 1
has x = - 2 and x = 2 as solutions.
Usually, we will write the solutions of an equation in set notation. This set is
called the solution set of the equation. For example, the solution set of the equation
x2 - 9 = 0 is 5 - 3, 36 .
Some equations have no real solution. For example, x2 + 9 = 5 has no real
solution, because there is no real number whose square, when added to 9, equals 5.
An equation that is satisfied for every value of the variable for which both sides
are defined is called an identity. For example, the equation
3x + 5 = x + 3 + 2x + 2
is an identity, because this statement is true for any real number x.
One method for solving an equation is to replace the original equation by a
succession of equivalent equations, equations having the same solution set,
until an equation with an obvious solution is obtained.
For example, consider the following succession of equivalent equations:
2x + 3 = 13
2x = 10
x = 5
We conclude that the solution set of the original equation is 5 56 .
How do we obtain equivalent equations? In general, there are five ways.
SECTION 1.1 Linear Equations
83
Procedures That Result in Equivalent Equations
1. Interchange the two sides of the equation:
Replace
3 = x by x = 3
2. Simplify the sides of the equation by combining like terms, eliminating
parentheses, and so on:
1x + 22 + 6 = 2x + 1x + 12
x + 8 = 3x + 1
Replace
by
3. Add or subtract the same expression on both sides of the equation:
3x - 5 = 4
13x - 52 + 5 = 4 + 5
Replace
by
4. Multiply or divide both sides of the equation by the same nonzero
expression:
3x
6
=
x ≠ 1
x - 1
x - 1
3x #
6 #
1x - 12 =
1x - 12
x - 1
x - 1
Replace
by
WARNING Squaring both sides of an
equation does not necessarily lead to
an equivalent equation. For example,
x = 3 has one solution, but x 2 = 9
has two solutions, x = - 3 and x = 3.
5. If one side of the equation is 0 and the other side can be factored, then we
may use the Zero-Product Property* and set each factor equal to 0:
x1x - 32 = 0
x = 0 or x - 3 = 0
Replace
by
■
Whenever it is possible to solve an equation in your head, do so. For example,
The solution of 2x = 8 is x = 4.
The solution of 3x - 15 = 0 is x = 5.
Now Work
EXAMPL E 1
PROBLEM
11
Solving an Equation
Solve the equation: 3x - 5 = 4
Solution
Replace the original equation by a succession of equivalent equations.
3x - 5 = 4
13x - 52 + 5 = 4 + 5
3x = 9
3x
9
=
3
3
x = 3
Add 5 to both sides.
Simplify.
Divide both sides by 3.
Simplify.
The last equation, x = 3, has the single solution 3. All these equations are equivalent,
so 3 is the only solution of the original equation, 3x - 5 = 4.
*The Zero-Product Property says that if ab = 0, then a = 0 or b = 0 or both equal 0.
84
CHAPTER 1 Equations and Inequalities
Check: Check the solution by substituting 3 for x in the original equation.
3x - 5 =
3132 - 5 ≟
9 - 5 ≟
4 =
4
4
4
4
The solution checks. The solution set is 5 36 .
Now Work
PROBLEM
r
25
Steps for Solving Equations
STEP 1:
STEP 2:
STEP 3:
STEP 4:
List any restrictions on the domain of the variable.
Simplify the equation by replacing the original equation by a succession
of equivalent equations using the procedures listed earlier.
If the result of Step 2 is a product of factors equal to 0, use the
Zero-Product Property and set each factor equal to 0 (procedure 5).
Check your solution(s).
1 Solve a Linear Equation
Linear equations are equations such as
3x + 12 = 0
DEFINITION
- 2x + 5 = 0
1
x - 23 = 0
2
A linear equation in one variable is an equation equivalent in form to
ax + b = 0
where a and b are real numbers and a ≠ 0.
Sometimes a linear equation is called a first-degree equation, because the left
side is a polynomial in x of degree 1.
It is relatively easy to solve a linear equation. The idea is to isolate the variable:
ax + b = 0
ax = - b
-b
x =
a
a ≠ 0
Subtract b from both sides.
Divide both sides by a, a ≠ 0.
The linear equation ax + b = 0, a ≠ 0, has the single solution given by the
b
formula x = - .
a
EX AMPLE 2
Solving a Linear Equation
Solve the equation:
1
1
1x + 52 - 4 = 12x - 12
2
3
SECTION 1.1 Linear Equations
Solution
85
To clear the equation of fractions, multiply both sides by 6, the least common
1
1
multiple (LCM) of the denominators of the fractions and .
2
3
1
1
1x + 52 - 4 = 12x - 12
2
3
1
1
6c 1x + 52 - 4 d = 6c 12x - 12 d Multiply both sides by 6, the LCM of 2 and 3.
2
3
31x + 52 - 6 # 4 = 212x - 12
3x + 15 - 24
3x - 9
3x - 9 + 9
3x
3x - 4x
-x
x
=
=
=
=
=
=
=
4x 4x 4x 4x +
4x +
7
-7
2
2
2 + 9
7
7 - 4x
Use the Distributive Property on the left and
the Associative Property on the right.
Use the Distributive Property.
Combine like terms.
Add 9 to each side.
Simplify.
Subtract 4x from each side.
Simplify.
Multiply both sides by - 1.
Check: Substitute - 7 for x in the expressions on the left and right sides of the
original equation, and simplify. If the two expressions are equal, the
solution checks.
1
1
1
1x + 52 - 4 = 1 - 7 + 52 - 4 = 1 - 22 - 4 = - 1 - 4 = - 5
2
2
2
1
1
1
1
12x - 12 = 3 21 - 72 - 14 = 1 - 14 - 12 = 1 - 152 = - 5
3
3
3
3
Since the two expressions are equal, the solution checks. The solution set is 5 - 76 .
Now Work
EXAMPL E 3
PROBLEM
r
35
Solving a Linear Equation Using a Calculator
Solve the equation: 2.78x +
2
= 54.06
17.931
Round the answer to two decimal places.
Solution
To avoid rounding errors, solve for x before using a calculator.
2.78x +
2
= 54.06
17.931
2
2
from each side.
Subtract
17.931
17.931
2
54.06 17.931
x =
Divide each side by 2.78.
2.78
2.78x = 54.06 -
Now use your calculator. The solution, rounded to two decimal places, is 19.41.
Check: Store the calculator solution 19.40592134 in memory, and proceed
2
to evaluate 2.78x +
.
17.931
12.782 119.405921342 +
Now Work
PROBLEM
67
2
= 54.06
17.931
r
86
CHAPTER 1 Equations and Inequalities
2 Solve Equations That Lead to Linear Equations
Solving an Equation That Leads to a Linear Equation
EX AMPL E 4
Solve the equation: 12y + 12 1y - 12 = 1y + 52 12y - 52
Solution
12y + 12 1y - 12
2y2 - y - 1
-y - 1
-y
- 6y
y
= 1y + 52 12y - 52
=
=
=
=
=
2y2 + 5y - 25
5y - 25
5y - 24
- 24
4
Multiply and combine like terms.
Subtract 2y2 from each side.
Add 1 to each side.
Subtract 5y from each side.
Divide both sides by - 6.
Check: 12y + 12 1y - 12 = 3 2142 + 14 14 - 12 = 18 + 12 132 = 192 132 = 27
1y + 52 12y - 52 = 14 + 52 3 2142 - 54 = 192 18 - 52 = 192 132 = 27
The two expressions are equal, so the solution checks. The solution set is 5 46 .
r
Solving an Equation That Leads to a Linear Equation
EX AMPLE 5
Solve the equation:
Solution
3
1
7
=
+
x - 2
x - 1
1x - 12 1x - 22
First, notice that the domain of the variable is 5 x x ≠ 1, x ≠ 26 . Clear the equation
of fractions by multiplying both sides by the least common multiple of the
denominators of the three fractions, 1x - 12 1x - 22.
3
1
7
=
+
x - 2
x - 1
1x - 12 1x - 22
1x - 12 1x - 22
3
1
7
= 1x - 12 1x - 22 c
+
d
x - 2
x - 1
1x - 12 1x - 22
3x - 3 = 1x - 12 1x - 22
1
7
+ 1x - 12 1x - 22
x - 1
1x - 12 1x - 22
3x - 3 = 1x - 22 + 7
3x - 3 = x + 5
Multiply both sides by
(x - 1)(x - 2); cancel
on the left.
Use the Distributive
Property on each side;
cancel on the right.
Simplify.
Combine like terms.
Add 3 to each side;
subtract x from each side.
2x = 8
x = 4
Divide by 2.
Check:
3
3
3
=
=
x - 2
4 - 2
2
1
7
1
7
1
7
2
7
9
3
+
=
+
= + # = + = =
x - 1
1x - 12 1x - 22
4 - 1
14 - 12 14 - 22
3
3 2
6
6
6
2
The solution checks. The solution set is 5 46 .
Now Work
EX AMPLE 6
PROBLEM
61
An Equation with No Solution
Solve the equation:
3x
3
+ 2 =
x - 1
x - 1
r
SECTION 1.1 Linear Equations
Solution
First, note that the domain of the variable is 5 x x ≠ 16 . Since the two quotients in
the equation have the same denominator, x - 1, simplify by multiplying both sides
by x - 1. The resulting equation is equivalent to the original equation, since we are
multiplying by x - 1, which is not 0. (Remember, x ≠ 1.)
3x
3
+ 2 =
x - 1
x - 1
3x
3
a
+ 2b # 1x - 12 =
x - 1
x - 1
3x
x - 1
NOTE Example 6 illustrates the
appearance of an extraneous solution.
This will be discussed more in
Section 1.4.
■
#
#
1x - 12
1x - 12 + 2 # 1x - 12 = 3
3x + 2x - 2
5x - 2
5x
x
=
=
=
=
Multiply both sides by
x - 1; cancel on the right.
Use the Distributive Property on
the left side; cancel on the left.
3
3
5
1
Simplify.
Combine like terms.
Add 2 to each side.
Divide both sides by 5.
The solution appears to be 1. But recall that x = 1 is not in the domain of the variable,
so this value must be discarded. The equation has no solution. The solution set is ∅.
r
Now Work
EXAMPL E 7
87
PROBLEM
51
Converting to Fahrenheit from Celsius
In the United States we measure temperature in both degrees Fahrenheit (°F) and
5
degrees Celsius (°C), which are related by the formula C = 1F - 322. What are the
9
Fahrenheit temperatures corresponding to Celsius temperatures of 0°, 10°, 20°, and 30°C?
Solution
We could solve four equations for F by replacing C each time by 0, 10, 20, and 30.
5
Instead, it is much easier and faster first to solve the equation C = 1F - 322
9
for F and then to substitute in the values of C.
5
1F - 322
9
9C = 51F - 322
C =
9C = 5F - 160
5F - 160 = 9C
5F = 9C + 160
9
F = C + 32
5
Multiply both sides by 9.
Use the Distributive Property.
Interchange sides.
Add 160 to each side.
Divide both sides by 5.
Now do the required arithmetic.
0°C: F =
9
102 + 32 = 32°F
5
9
1102 + 32 = 50°F
5
9
20°C: F = 1202 + 32 = 68°F
5
9
30°C: F = 1302 + 32 = 86°F
5
10°C: F =
NOTE The icon icon is a Model It! icon. It
indicates that the discussion or problem
involves modeling.
■
r
3 Solve Problems That Can Be Modeled by Linear Equations
Although each situation has its unique features, we can provide an outline of the
steps to follow when solving applied problems.
88
CHAPTER 1 Equations and Inequalities
Steps for Solving Applied Problems
STEP 1:
NOTE It is a good practice to choose
a variable that reminds you of the
unknown. For example, use t for time.
■
STEP 2:
STEP 3:
STEP 4:
STEP 5:
EX AMPLE 8
Read the problem carefully, perhaps two or three times. Pay particular
attention to the question being asked in order to identify what you
are looking for. Identify any relevant formulas you may need (d = rt,
A = pr 2, etc.). If you can, determine realistic possibilities for the answer.
Assign a letter (variable) to represent what you are looking for, and, if
necessary, express any remaining unknown quantities in terms of this
variable.
Make a list of all the known facts, and translate them into mathematical
expressions. These may take the form of an equation (or, later, an
inequality) involving the variable. The equation (or inequality) is
called the model. If possible, draw an appropriately labeled diagram
to assist you. Sometimes a table or chart helps.
Solve the equation for the variable, and then answer the question,
usually using a complete sentence.
Check the answer with the facts in the problem. If it agrees,
congratulations! If it does not agree, try again.
Investments
A total of $18,000 is invested, some in stocks and some in bonds. If the amount invested
in bonds is half that invested in stocks, how much is invested in each category?
Step-by-Step Solution
Step 1: Determine what you are
looking for.
We are being asked to find the amount of two investments. These amounts must total
$18,000. (Do you see why?)
Step 2: Assign a variable to
represent what you are looking for.
If necessary, express any remaining
unknown quantities in terms of this
variable.
If x equals the amount invested in stocks, then the rest of the money, 18,000 - x, is
the amount invested in bonds.
Step 3: Translate the English into
mathematical statements. It may
be helpful to draw a figure that
represents the situation. Sometimes
a table can be used to organize the
information. Use the information to
build your model.
Set up a table:
Step 4: Solve the equation and
answer the original question.
Amount in Stocks
Amount in Bonds
Reason
x
18,000 - x
Total invested is $18,000.
We also know that:
Total amount invested in bonds
is
18,000 - x
=
18,000 - x =
one@half that in stocks
1
1x2
2
1
x
2
18,000 = x +
1
x
2
Add x to both sides.
3
x
Simplify.
2
2
2 3
2
a b 18,000 = a b a xb Multiply both sides by .
3
3
3 2
12,000 = x
Simplify.
So $12,000 is invested in stocks, and $18,000 - $12,000 = $6000 is invested in bonds.
18,000 =
Step 5: Check your answer with the
facts presented in the problem.
The total invested is $12,000 + $6000 = $18,000, and the amount in bonds, $6000,
is half that in stocks, $12,000.
Now Work
PROBLEM
83
r
SECTION 1.1 Linear Equations
89
Determining an Hourly Wage
EXAMPL E 9
Shannon grossed $725 one week by working 52 hours. Her employer pays
time-and-a-half for all hours worked in excess of 40 hours. With this information, can
you determine Shannon’s regular hourly wage?
Solution
STEP 1: We are looking for an hourly wage. Our answer will be expressed in dollars
per hour.
STEP 2: Let x represent the regular hourly wage, measured in dollars per hour. Then
1.5x is the overtime hourly wage.
STEP 3: Set up a table:
Hours Worked
Hourly Wage
Salary
Regular
40
x
40x
Overtime
12
1.5x
12(1.5x) = 18x
The sum of regular salary plus overtime salary will equal $725. From the
table, 40x + 18x = 725.
STEP 4:
40x + 18x = 725
58x = 725
x = 12.50
Shannon’s regular hourly wage is $12.50 per hour.
STEP 5: Forty hours yields a salary of 40 112.502 = $500, and 12 hours of overtime
yields a salary of 1211.52 112.502 = $225, for a total of $725.
Now Work
PROBLEM
85
r
SUMMARY
Steps for Solving a Linear Equation
To solve a linear equation, follow these steps:
STEP 1: List any restrictions on the variable.
STEP 2: If necessary, clear the equation of fractions by multiplying both sides by the least common multiple (LCM)
of the denominators of all the fractions.
STEP 3: Remove all parentheses and simplify.
STEP 4: Collect all terms containing the variable on one side and all remaining terms on the other side.
STEP 5: Simplify and solve.
STEP 6: Check your solution(s).
Historical Feature
S
olving equations is among the oldest of mathematical activities,
and efforts to systematize this activity determined much of the
shape of modern mathematics.
Consider the following problem and its solution using only words:
Solve the problem of how many apples Jim has, given that
“Bob’s five apples and Jim’s apples together make twelve apples”
by thinking,
“Jim’s apples are all twelve apples less Bob’s five apples” and then
concluding,
“Jim has seven apples.”
The mental steps translated into algebra are
5 + x = 12
x = 12 - 5
= 7
The solution of this problem using only words is the earliest form
of algebra. Such problems were solved exactly this way in Babylonia in
1800 BC. We know almost nothing of mathematical work before this date,
although most authorities believe the sophistication of the earliest known
texts indicates that there must have been a long period of previous
development. The method of writing out equations in words persisted for
thousands of years, and although it now seems extremely cumbersome,
it was used very effectively by many generations of mathematicians.
The Arabs developed a good deal of the theory of cubic equations while
writing out all the equations in words. About AD 1500, the tendency to
abbreviate words in the written equations began to lead in the direction
of modern notation; for example, the Latin word et (meaning “and”)
developed into the plus sign, + . Although the occasional use of letters
to represent variables dates back to AD 1200, the practice did not become
common until about AD 1600. Development thereafter was rapid, and by
1635 algebraic notation did not differ essentially from what we use now.
90
CHAPTER 1 Equations and Inequalities
1.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
x
3. The domain of the variable in the expression
is
x - 4
. (p. 21)
1. The fact that 21x + 32 = 2x + 6 is attributable to the
Property. (pp. 9–13)
2. The fact that 3x = 0 implies that x = 0 is a result of the
Property. (pp. 9–13)
Concepts and Vocabulary
8. True or False Some equations have no solution.
4. True or False Multiplying both sides of an equation by any
number results in an equivalent equation.
9. An admissible value for the variable that makes the equation
of the equation.
a true statement is called a(n)
(a) identity
(b) solution
(c) degree
(d) model
5. An equation that is satisfied for every value of the variable
for which both sides are defined is called a(n)
.
6. An equation of the form ax + b = 0 is called a(n)
equation or a(n)
equation.
10. A chemist mixes 10 liters of a 20% solution with x liters of a
35% solution. Which of the following expressions represents
the total number of liters in the mixture?
35
(a) x
(b) 20 - x
(c)
(d) 10 + x
x
7. True or False The solution of the equation 3x - 8 = 0
3
is .
8
Skill Building
In Problems 11–18, mentally solve each equation.
11. 7x = 21
12. 6x = - 24
13. 3x + 15 = 0
15. 2x - 3 = 0
16. 3x + 4 = 0
17.
14. 6x + 18 = 0
1
5
x =
3
12
18.
2
9
x =
3
2
In Problems 19–66, solve each equation.
19. 3x + 4 = x
20. 2x + 9 = 5x
21. 2t - 6 = 3 - t
22. 5y + 6 = - 18 - y
23. 6 - x = 2x + 9
24. 3 - 2x = 2 - x
25. 3 + 2n = 4n + 7
26. 6 - 2m = 3m + 1
27. 213 + 2x2 = 31x - 42
28. 312 - x2 = 2x - 1
29. 8x - 13x + 22 = 3x - 10
30. 7 - 12x - 12 = 10
31.
1
1
3
x + 2 = - x
2
2
2
34. 1 -
1
x = 6
2
37. 0.9t = 0.4 + 0.1t
32.
2
1
x = 2 - x
3
3
33.
1
3
x - 5 = x
2
4
35.
2
1
1
p = p +
3
2
3
36.
1
1
4
- p =
2
3
3
39.
x + 2
x + 1
+
= 2
3
7
4
5
- 5 =
y
2y
38. 0.9t = 1 + t
40.
2x + 1
+ 16 = 3x
3
41.
4
2
+ = 3
y
y
42.
43.
1
2
3
+ =
2
x
4
44.
3
1
1
- =
x
3
6
45. 1x + 72 1x - 12 = 1x + 12 2
46. 1x + 22 1x - 32 = 1x + 32 2
47. x12x - 32 = 12x + 12 1x - 42
48. x11 + 2x2 = 12x - 12 1x - 22
49. z1z2 + 12 = 3 + z3
50. w 14 - w 2 2 = 8 - w 3
51.
2
x
+ 3 =
x - 2
x - 2
52.
2x
-6
=
- 2
x + 3
x + 3
53.
4
3
2x
= 2
x
+
2
x - 4
x - 4
54.
4
x
3
+
= 2
x
+
3
x - 9
x - 9
55.
3
x
=
x + 2
2
56.
3x
= 2
x - 1
57.
5
3
=
2x - 3
x + 5
58.
-3
-4
=
x + 4
x + 6
59.
6t + 7
3t + 8
=
4t - 1
2t - 4
60.
8w + 5
4w - 3
=
10w - 7
5w + 7
2
2
SECTION 1.1 Linear Equations
91
61.
-3
7
4
=
+
x - 2
x + 5
1x + 52 1x - 22
62.
-4
1
1
+
=
2x + 3
x-1
12x + 32 1x - 12
63.
2
3
5
+
=
y + 3
y - 4
y + 6
64.
4
-3
5
+
=
5z - 11
2z - 3
5 - z
65.
x + 3
-3
x
- 2
= 2
x2 - 1
x - x
x + x
66.
x + 1
x + 4
-3
- 2
= 2
x2 + 2x
x + x
x + 3x + 2
In Problems 67–70, use a calculator to solve each equation. Round the solution to two decimal places.
67. 3.2x +
21.3
= 19.23
65.871
69. 14.72 - 21.58x =
68. 6.2x -
18
x + 2.4
2.11
19.1
= 0.195
83.72
70. 18.63x -
14
21.2
=
x - 20
2.6
2.32
Applications and Extensions
In Problems 71–74, solve each equation. The letters a, b, and c are constants.
71. ax - b = c, a ≠ 0
73.
72. 1 - ax = b, a ≠ 0
x
x
+
= c, a ≠ 0, b ≠ 0, a ≠ - b
a
b
75. Find the number a for which x = 4 is a solution of the
equation
x + 2a = 16 + ax - 6a
74.
a
b
+
= c, c ≠ 0
x
x
76. Find the number b for which x = 2 is a solution of the
equation
x + 2b = x - 4 + 2bx
Problems 77–82 list some formulas that occur in applications. Solve each formula for the indicated variable.
77. Electricity
1
1
1
=
+
R
R1
R2
78. Finance A = P 11 + rt2
for R
for r
2
mv
for R
R
80. Chemistry PV = nRT for T
a
81. Mathematics S =
for r
1 - r
82. Mechanics v = - gt + v0 for t
79. Mechanics F =
83. Finance A total of $20,000 is to be invested, some in bonds
and some in certificates of deposit (CDs). If the amount
invested in bonds is to exceed that in CDs by $3000, how
much will be invested in each type of investment?
84. Finance A total of $10,000 is to be divided between Sean
and George, with George to receive $3000 less than Sean.
How much will each receive?
85. Computing Hourly Wages
Sandra, who is paid
time-and-a-half for hours worked in excess of 40 hours, had
gross weekly wages of $546 for 48 hours worked. What is her
regular hourly rate?
86. Computing Hourly Wages Leigh is paid time-and-a-half
for hours worked in excess of 40 hours and double-time for
hours worked on Sunday. If Leigh had gross weekly wages of
$627 for working 50 hours, 4 of which were on Sunday, what
is her regular hourly rate?
87. Computing Grades Going into the final exam, which will
count as two tests, Brooke has test scores of 80, 83, 71, 61,
and 95. What score does Brooke need on the final in order
to have an average score of 80?
88. Computing Grades Going into the final exam, which will
count as two-thirds of the final grade, Mike has test scores
of 86, 80, 84, and 90. What minimum score does Mike need
on the final in order to earn a B, which requires an average
score of 80? What does he need to earn an A, which requires
an average of 90?
89. Business: Discount Pricing A builder of tract homes reduced
the price of a model by 15%. If the new price is $425,000,
what was its original price? How much can be saved by
purchasing the model?
90. Business: Discount Pricing A car dealer, at a year-end
clearance, reduces the list price of last year’s models by 15%.
If a certain four-door model has a discounted price of $8000,
what was its list price? How much can be saved by purchasing
last year’s model?
91. Personal Finance: Concession Markup A movie theater
marks up the candy it sells by 275%. If a box of candy sells
for $3.00 at the theater, how much did the theater pay for
the box?
92
CHAPTER 1 Equations and Inequalities
92. Personal Finance: Cost of a Car The suggested list price of
a new car is $18,000. The dealer’s cost is 85% of list. How
much will you pay if the dealer is willing to accept $100 over
cost for the car?
93. Business: Theater Attendance The manager of the Coral
Theater wants to know whether the majority of its patrons
are adults or children. One day in July, 5200 tickets were
sold and the receipts totaled $29,961. The adult admission is
$7.50, and the children’s admission is $4.50. How many adult
patrons were there?
94. Business: Discount Pricing A wool suit, discounted by 30%
for a clearance sale, has a price tag of $399. What was the
suit’s original price?
95. Geometry The perimeter of a rectangle is 60 feet. Find its
length and width if the length is 8 feet longer than the width.
Tyshira tracks her net calories
98. Counting Calories
(calories taken in minus calories burned) as part of her
fitness program. For one particular day, her net intake was
1480 calories. Her lunch calories were half her breakfast
calories, and her dinner calories were 200 more than her
breakfast calories. She ate 120 less calories in snacks than
for breakfast, and she burned 700 calories by exercising
on her elliptical. How many calories did she take in from
snacks?
99. Sharing the Cost of a Pizza Judy and Tom agree to share the
cost of an $18 pizza based on how much each ate. If Tom
2
ate the amount that Judy ate, how much should each pay?
3
[Hint: Some pizza may be left.]
Tom’s portion
96. Geometry The perimeter of a rectangle is 42 meters. Find its
length and width if the length is twice the width.
97. Counting Calories
Herschel uses an app on his
smartphone to keep track of his daily calories from meals.
One day his calories from breakfast were 125 more than his
calories from lunch, and his calories from dinner were 300
less than twice his calories from lunch. If his total caloric
intake from meals was 2025, determine his calories for each
meal.
Judy’s portion
Explaining Concepts: Discussion and Writing
100. What Is Wrong? One step in the following list contains an
error. Identify it and explain what is wrong.
x = 2
3x - 2x = 2
3x = 2x + 2
x2 + 3x = x2 + 2x + 2
x + 3x - 10 = x2 + 2x - 8
2
1x - 22 1x + 52 = 1x - 22 1x + 42
x + 5 = x + 4
1 = 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
101. The equation
8 + x
5
+ 3 =
x + 3
x + 3
has no solution, yet when we go through the process of
solving it, we obtain x = - 3. Write a brief paragraph to
explain what causes this to happen.
102. Make up an equation that has no solution and give it to a
fellow student to solve. Ask the fellow student to write a
critique of your equation.
‘Are You Prepared?’ Answers
1. Distributive
2. Zero-Product
3. 5x x ≠ 46
1.2 Quadratic Equations
PREPARING FOR THIS SECTION Before getting started, review the following:
r Factoring Polynomials (Section R.5, pp. 49–55)
r Zero-Product Property (Section R.1, p. 13)
r Square Roots (Section R.2, pp. 23–24)
r Complete the Square (Section R.5, p. 56)
Now Work the ‘Are You Prepared?’ problems on page 101.
OBJECTIVES 1
2
3
4
Solve a Quadratic Equation by Factoring (p. 93)
Solve a Quadratic Equation by Completing the Square (p. 95)
Solve a Quadratic Equation Using the Quadratic Formula (p. 96)
Solve Problems That Can Be Modeled by Quadratic Equations (p. 99)
SECTION 1.2 Quadratic Equations
93
Quadratic equations are equations such as
2x2 + x + 8 = 0
3x2 - 5x + 6 = 0
x2 - 9 = 0
DEFINITION
A quadratic equation is an equation equivalent to one of the form
ax2 + bx + c = 0
(1)
where a, b, and c are real numbers and a ≠ 0.
A quadratic equation written in the form ax2 + bx + c = 0 is said to be in
standard form.
Sometimes, a quadratic equation is called a second-degree equation because,
when it is in standard form, the left side is a polynomial of degree 2. We shall discuss
three ways of solving quadratic equations: by factoring, by completing the square,
and by using the quadratic formula.
1 Solve a Quadratic Equation by Factoring
When a quadratic equation is written in standard form, it may be possible to
factor the expression on the left side into the product of two first-degree
polynomials. Then, by using the Zero-Product Property and setting each factor
equal to 0, the resulting linear equations can be solved to obtain the solutions of the
quadratic equation.
EXAMPL E 1
Solving a Quadratic Equation by Factoring
Solve the equations:
(a) x2 + 6x = 0
Solution
(b) 2x2 = x + 3
(a) The equation is in the standard form specified in equation (1). The left side may
be factored as
x2 + 6x = 0
x(x + 6) = 0 Factor.
Using the Zero-Product Property, set each factor equal to 0 and then solve
the resulting first-degree equations.
x = 0
x = 0
or
or
The solution set is 5 0, - 66 .
x + 6 = 0
x = -6
Zero-Product Property
Solve.
(b) Place the equation 2x2 = x + 3 in standard form by subtracting x and 3 from
both sides.
2x2 = x + 3
2
2x - x - 3 = 0
Subtract x and 3 from both sides.
The left side may now be factored as
12x - 32 1x + 12 = 0
Factor.
so that
2x - 3 = 0 or
3
x =
2
3
The solution set is e - 1, f.
2
x + 1 = 0
x = -1
Zero-Product Property
Solve.
r
94
CHAPTER 1 Equations and Inequalities
When the left side factors into two linear equations with the same solution, the
quadratic equation is said to have a repeated solution. This solution is also called a
root of multiplicity 2, or a double root.
EX AMPLE 2
Solving a Quadratic Equation by Factoring
Solve the equation: 9x2 - 6x + 1 = 0
Solution
This equation is already in standard form, and the left side can be factored.
9x2 - 6x + 1 = 0
13x - 12 13x - 12 = 0
so
x =
1
3
or x =
1
3
1
1
This equation has only the repeated solution . The solution set is e f.
3
3
Now Work
PROBLEMS
13
AND
r
23
The Square Root Method
Suppose that we wish to solve the quadratic equation
x2 = p
(2)
where p Ú 0 is a nonnegative number. Proceeding as in the earlier examples,
x2 - p = 0
Put in standard form.
1x - 1p2 1x + 1p2 = 0
Factor (over the real numbers).
x = 1p or x = - 1p Solve.
We have the following result:
If x2 = p and p Ú 0, then x = 1p or x = - 1p .
(3)
When statement (3) is used, it is called the Square Root Method. In
statement (3), note that if p 7 0 the equation x2 = p has two solutions, x = 1p
and x = - 1p . We usually abbreviate these solutions as x = { 1p , which is
read as “x equals plus or minus the square root of p.”
For example, the two solutions of the equation
x2 = 4
are
x = { 24
Use the Square Root Method.
and, since 24 = 2, we have
x = {2
The solution set is 5 - 2, 26 .
EX AMPLE 3
Solving a Quadratic Equation Using the Square Root Method
Solve each equation.
(a) x2 = 5
Solution
(b) 1x - 22 2 = 16
(a) Use the Square Root Method to get
x2 = 5
x = { 25
x = 25 or x = - 25
The solution set is 5 - 15, 156 .
Use the Square Root Method.
SECTION 1.2 Quadratic Equations
95
(b) Use the Square Root Method to get
1x - 22 2 = 16
x - 2 =
x - 2 =
x - 2 =
x =
{ 216
Use the Square Root Method.
{4
116 = 4
4 or x - 2 = - 4
6 or
x = -2
The solution set is 5 - 2, 66 .
Now Work
PROBLEM
r
33
2 Solve a Quadratic Equation by Completing the Square
EXAMPL E 4
Solving a Quadratic Equation by Completing the Square
Solve by completing the square: x2 + 5x + 4 = 0
Solution
Always begin this procedure by rearranging the equation so that the constant is on
the right side.
x2 + 5x + 4 = 0
x2 + 5x = - 4
NOTE If the coefficient of the square
term is not 1, divide both sides by the
coefficient of the square term before
attempting to complete the square. ■
Since the coefficient of x2 is 1, complete the square on the left side by adding
2
1
25
a # 5b =
. Of course, in an equation, whatever is added to the left side must
2
4
25
also be added to the right side. So add
to both sides.
4
25
25
25
x2 + 5x +
= -4 +
Add
to both sides.
4
4
4
5 2
9
ax + b =
Factor the left side.
2
4
5
9
x + = {
Use the Square Root Method.
74
2
3
9
5
19
3
x + = {
=
=
2
2
A4
2
14
5
3
x = - {
2
2
5
3
5
3
x = - + = - 1 or x = - - = - 4
2
2
2
2
The solution set is 5 - 4, - 16 .
T H E S O LU T I O N O F T H E E Q UAT I O N I N E X A M P L E
r
4
B E O BTA I N E D BY FAC TO R I N G . R E W O R K E X A M P L E
FAC TO R I N G .
CAN ALSO
4
USING
The next example illustrates an equation that cannot be solved by factoring.
EXAMPL E 5
Solving a Quadratic Equation by Completing the Square
Solve by completing the square: 2x2 - 8x - 5 = 0
Solution
First, rewrite the equation so that the constant is on the right side.
2x2 - 8x - 5 = 0
2x2 - 8x = 5 Add 5 to both sides.
96
CHAPTER 1 Equations and Inequalities
Next, divide both sides by 2 so that the coefficient of x2 is 1. (This enables us to
complete the square at the next step.)
x2 - 4x =
5
2
Finally, complete the square by adding 4 to both sides.
5
x2 - 4x + 4 = + 4
Add 4 to both sides.
2
13
1x - 22 2 =
Factor on the left; simplify on the right.
2
13
x - 2 = {
Use the Square Root Method.
A2
x - 2 = {
226
2
x = 2 {
NOTE If we wanted an approximation,
say rounded to two decimal places, of
these solutions, we would use a calculator
to get { - 0.55, 4.55}.
■
The solution set is e 2 -
Now Work
13
113
113
=
=
A 2
12
12
# 12
12
=
126
2
226
2
226
226
,2 +
f.
2
2
PROBLEM
r
37
3 Solve a Quadratic Equation Using the Quadratic Formula
We can use the method of completing the square to obtain a general formula for
solving any quadratic equation
ax2 + bx + c = 0
NOTE There is no loss in generality
to assume that a 7 0, since if a 6 0
we can multiply by - 1 to obtain an
equivalent equation with a positive
leading coefficient.
■
a ≠ 0
As in Examples 4 and 5, rearrange the terms as
ax2 + bx = - c
a 7 0
Since a 7 0, divide both sides by a to get
x2 +
c
b
x = a
a
Now the coefficient of x2 is 1. To complete the square on the left side, add the square
1
of of the coefficient of x; that is, add
2
1 b 2
b2
a # b =
2 a
4a2
to both sides. Then
x2 +
b
b2
b2
c
x +
=
2
a
a
4a
4a2
ax +
b2 - 4ac
b 2
b =
2a
4a2
b2
b2
c
4ac
b 2 - 4ac
- =
=
2
2
2
a
4a
4a
4a
4a 2
Provided that b2 - 4ac Ú 0, we can now use the Square Root Method to get
x +
b2 - 4ac
b
= {
2a
B 4a2
b
{ 2b2 - 4ac
x +
=
2a
2a
The square root of a quotient equals
the quotient of the square roots.
Also, 24a2 = 2a since a 7 0.
(4)
SECTION 1.2 Quadratic Equations
x = =
b
2b2 - 4ac
{
2a
2a
- b { 2b2 - 4ac
2a
Add -
97
b
to both sides.
2a
Combine the quotients on the right.
What if b2 - 4ac is negative? Then equation (4) states that the left expression
(a real number squared) equals the right expression (a negative number). Since this
is impossible for real numbers, we conclude that if b2 - 4ac 6 0, the quadratic
equation has no real solution. (We discuss quadratic equations for which the
quantity b2 - 4ac 6 0 in detail in the next section.)
THEOREM
Quadratic Formula
Consider the quadratic equation
ax2 + bx + c = 0
a ≠ 0
If b2 - 4ac 6 0, this equation has no real solution.
If b2 - 4ac Ú 0, the real solution(s) of this equation is (are) given by the
quadratic formula:
x =
- b { 2b2 - 4ac
2a
(5)
The quantity b2 − 4ac is called the discriminant of the quadratic equation, because
its value tells us whether the equation has real solutions. In fact, it also tells us how many
solutions to expect.
Discriminant of a Quadratic Equation
For a quadratic equation ax2 + bx + c = 0, a ≠ 0:
1. If b2 - 4ac 7 0, there are two unequal real solutions.
2. If b2 - 4ac = 0, there is a repeated solution, a double root.
3. If b2 - 4ac 6 0, there is no real solution.
When asked to find the real solutions, of a quadratic equation, always evaluate
the discriminant first to see if there are any real solutions.
EXAMPL E 6
Solving a Quadratic Equation Using the Quadratic Formula
Use the quadratic formula to find the real solutions, if any, of the equation
3x2 - 5x + 1 = 0
Solution
The equation is in standard form, so compare it to ax2 + bx + c = 0 to find a, b,
and c.
3x2 - 5x + 1 = 0
ax2 + bx + c = 0 a = 3, b = - 5, c = 1
98
CHAPTER 1 Equations and Inequalities
With a = 3, b = - 5, and c = 1, evaluate the discriminant b2 - 4ac.
b2 - 4ac = 1 - 52 2 - 4132 112 = 25 - 12 = 13
Since b2 - 4ac 7 0, there are two real solutions, which can be found using the
quadratic formula.
x =
- 1 - 52 { 213
- b { 2b2 - 4ac
5 { 213
=
=
2a
2132
6
The solution set is e
EX AMPLE 7
5 - 213 5 + 213
,
f.
6
6
r
Solving a Quadratic Equation Using the Quadratic Formula
Use the quadratic formula to find the real solutions, if any, of the equation
25 2
x - 30x + 18 = 0
2
Solution
The equation is given in standard form. However, to simplify the arithmetic, clear
the fraction.
25 2
x - 30x + 18 = 0
2
25x2 - 60x + 36 = 0
Clear fraction; multiply by 2.
ax2 + bx + c = 0
Compare to standard form.
With a = 25, b = - 60, and c = 36, evaluate the discriminant.
b2 - 4ac = 1 - 602 2 - 41252 1362 = 3600 - 3600 = 0
The equation has a repeated solution, which is found by using the quadratic formula.
60 { 20
60
6
- b { 2b2 - 4ac
=
=
=
2a
50
50
5
x =
6
The solution set is e f.
5
EX AMPLE 8
r
Solving a Quadratic Equation Using the Quadratic Formula
Use the quadratic formula to find the real solutions, if any, of the equation
3x2 + 2 = 4x
Solution
The equation, as given, is not in standard form.
3x2 + 2 = 4x
3x2 - 4x + 2 = 0
Put in standard form.
ax + bx + c = 0
Compare to standard form.
2
With a = 3, b = - 4, and c = 2, the discriminant is
b2 - 4ac = 1 - 42 2 - 4132 122 = 16 - 24 = - 8
Since b2 - 4ac 6 0, the equation has no real solution.
Now Work
PROBLEMS
47
AND
57
r
SECTION 1.2 Quadratic Equations
EXAMPL E 9
Solving a Quadratic Equation Using the Quadratic Formula
Find the real solutions, if any, of the equation: 9 +
Solution
99
2
3
- 2 = 0, x ≠ 0
x
x
In its present form, the equation
9 +
3
2
- 2 = 0
x
x
is not a quadratic equation. However, it can be transformed into one by multiplying
each side by x2. The result is
9x2 + 3x - 2 = 0
Although we multiplied each side by x2, we know that x2 ≠ 0 (do you see why?), so
this quadratic equation is equivalent to the original equation.
Using a = 9, b = 3, and c = - 2, the discriminant is
b2 - 4ac = 32 - 4192 1 - 22 = 9 + 72 = 81
Since b2 - 4ac 7 0, the new equation has two real solutions.
x =
- b { 2b2 - 4ac
- 3 { 281
-3 { 9
=
=
2a
2192
18
x =
-3 + 9
6
1
=
=
18
18
3
or x =
-3 - 9
- 12
2
=
= 18
18
3
2 1
The solution set is e - , f.
3 3
r
SUMMARY
Procedure for Solving a Quadratic Equation
To solve a quadratic equation, first put it in standard form:
ax2 + bx + c = 0
Then:
STEP 1: Identify a, b, and c.
STEP 2: Evaluate the discriminant, b2 - 4ac.
STEP 3: (a) If the discriminant is negative, the equation has no real solution.
(b) If the discriminant is zero, the equation has one real solution, a double root.
(c) If the discriminant is positive, the equation has two distinct real solutions.
If you can easily spot factors, use the factoring method to solve the equation. Otherwise, use the quadratic formula
or the method of completing the square.
4 Solve Problems That Can Be Modeled by Quadratic Equations
Many applied problems require the solution of a quadratic equation. Let’s look at
one that you will probably see again in a slightly different form if you study calculus.
EX AM PL E 1 0
Constructing a Box
From each corner of a square piece of sheet metal, remove a square of side 9 centimeters.
Turn up the edges to form an open box. If the box is to hold 144 cubic centimeters
(cm3), what should be the dimensions of the piece of sheet metal?
100
CHAPTER 1 Equations and Inequalities
Solution
Use Figure 1 as a guide. We have labeled by x the length of a side of the square
piece of sheet metal. The box will be of height 9 centimeters, and its square base will
measure x - 18 on each side. The volume V 1Length * Width * Height2 of the
box is therefore
V = 1x - 182 1x - 182 # 9 = 91x - 182 2
x cm
9 cm
9 cm
9 cm
9 cm
x 18
9 cm
9 cm
x 18
9 cm
x cm
x 18
9 cm
x 18
Volume 9(x 18)(x 18)
9 cm
Figure 1
Since the volume of the box is to be 144 cm3, we have
Check: If we take a piece of
sheet metal 22 centimeters by 22
centimeters, cut out a 9-centimeter
square from each corner, and fold
up the edges, we get a box whose
dimensions are 9 by 4 by 4, with
volume 9 * 4 * 4 = 144 cm3, as
required.
91x - 182 2
1x - 182 2
x - 18
x
x
=
=
=
=
=
144
V = 144
16
Divide each side by 9.
{4
Use the Square Root Method.
18 { 4
22 or x = 14
Discard the solution x = 14 (do you see why?) and conclude that the sheet metal
should be 22 centimeters by 22 centimeters.
r
Now Work
PROBLEM
99
Historical Feature
P
roblems using quadratic equations are found in the oldest
known mathematical literature. Babylonians and Egyptians
were solving such problems before 1800 BC. Euclid solved
quadratic equations geometrically in his Data (300 BC), and the Hindus
and Arabs gave rules for solving any quadratic equation with real roots.
Because negative numbers were not freely used before AD 1500, there
were several different types of quadratic equations, each with its own
rule. Thomas Harriot (1560–1621) introduced the method of factoring to
obtain solutions, and François Viète (1540–1603) introduced a method
that is essentially completing the square.
Until modern times it was usual to neglect the negative roots (if there
were any), and equations involving square roots of negative quantities
were regarded as unsolvable until the 1500s.
Historical Problems
1. One of al-Khwǎrízmí solutions Solve x2 + 12x = 85 by drawing
the square shown. The area of the four white rectangles and the
yellow square is x2 + 12x. We then set this expression equal to
85 to get the equation x2 + 12x = 85. If we add the four blue
x
3
x
Area = x 2
3
Area = 3x
u2 + (2z + 12)u + (z2 + 12z - 85) = 0
3
x
x
3
3
3
2. Viète’s method Solve x2 + 12x - 85 = 0 by letting x = u + z.
Then
(u + z)2 + 12(u + z) - 85 = 0
3
3
squares, we will have a larger square of known area. Complete the
solution.
Now select z so that 2z + 12 = 0 and finish the solution.
3. Another method to get the quadratic formula Look at equation (4)
2b2 - 4ac 2
on page 96. Rewrite the right side as a
b and then
2a
subtract it from each side. The right side is now 0 and the left side is a
difference of two squares. If you factor this difference of two squares,
you will easily be able to get the quadratic formula, and moreover,
the quadratic expression is factored, which is sometimes useful.
SECTION 1.2 Quadratic Equations
101
1.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
4. True or False 2x2 = 0 x 0 . (pp. 23–24)
1. Factor: x2 - 5x - 6 (pp. 49–55)
2
2. Factor: 2x - x - 3 (pp. 49–55)
5. Complete the square of x2 + 5x. Factor the new
expression. (p. 56)
3. The solution set of the equation 1x - 32 13x + 52 = 0
is
. (p. 13)
Concepts and Vocabulary
9. A quadratic equation is sometimes called a
(a) first-degree
(b) second-degree
(c) third-degree
(d) fourth-degree
6. The quantity b2 - 4ac is called the
of a
quadratic equation. If it is
, the equation has no real
solution.
equation.
10. Which of the following quadratic equations is in standard
form?
(a) x2 - 7x = 5
(b) 9 = x2
(c) (x + 5)(x - 4) = 0
(d) 0 = 5x2 - 6x - 1
7. True or False Quadratic equations always have two real
solutions.
8. True or False If the discriminant of a quadratic equation
is positive, then the equation has two solutions that are
negatives of one another.
Skill Building
In Problems 11–30, solve each equation by factoring.
11. x2 - 9x = 0
12. x2 + 4x = 0
13. x2 - 25 = 0
14. x2 - 9 = 0
15. z2 + z - 6 = 0
16. v2 + 7v + 6 = 0
17. 2x2 - 5x - 3 = 0
18. 3x2 + 5x + 2 = 0
19. 3t 2 - 48 = 0
20. 2y2 - 50 = 0
21. x1x - 82 + 12 = 0
22. x1x + 42 = 12
23. 4x2 + 9 = 12x
24. 25x2 + 16 = 40x
25. 61p2 - 12 = 5p
26. 212u2 - 4u2 + 3 = 0
27. 6x - 5 =
6
x
28. x +
12
= 7
x
29.
In Problems 31–36, solve each equation by the Square Root Method.
41x - 22
x - 3
+
3
-3
=
x
x1x - 32
30.
5
3
= 4 +
x + 4
x - 2
31. x2 = 25
32. x2 = 36
33. 1x - 12 2 = 4
34. 1x + 22 2 = 1
35. 12y + 32 2 = 9
36. 13z - 22 2 = 4
In Problems 37–42, solve each equation by completing the square.
37. x2 + 4x = 21
40. x2 +
2
1
x - = 0
3
3
38. x2 - 6x = 13
41. 3x2 + x -
1
= 0
2
39. x2 -
1
3
x = 0
2
16
42. 2x2 - 3x - 1 = 0
In Problems 43–66, find the real solutions, if any, of each equation. Use the quadratic formula.
43. x2 - 4x + 2 = 0
44. x2 + 4x + 2 = 0
45. x2 - 4x - 1 = 0
46. x2 + 6x + 1 = 0
47. 2x2 - 5x + 3 = 0
48. 2x2 + 5x + 3 = 0
49. 4y2 - y + 2 = 0
50. 4t 2 + t + 1 = 0
51. 4x2 = 1 - 2x
52. 2x2 = 1 - 2x
53. 4x2 = 9x
54. 5x = 4x2
55. 9t 2 - 6t + 1 = 0
56. 4u2 - 6u + 9 = 0
57.
3 2
1
1
x - x - = 0
4
4
2
102
58.
CHAPTER 1 Equations and Inequalities
2 2
x - x - 3 = 0
3
61. 2x1x + 22 = 3
64. 4 +
1
1
- 2 = 0
x
x
59.
5 2
1
x - x =
3
3
60.
62. 3x1x + 22 = 1
65.
1
3 2
x - x =
5
5
63. 4 -
1
3x
+ = 4
x - 2
x
66.
2
1
- 2 = 0
x
x
2x
1
+ = 4
x - 3
x
In Problems 67–72, find the real solutions, if any, of each equation. Use the quadratic formula and a calculator. Express any solutions
rounded to two decimal places.
67. x2 - 4.1x + 2.2 = 0
68. x2 + 3.9x + 1.8 = 0
69. x2 + 23 x - 3 = 0
70. x2 + 22 x - 2 = 0
71. px2 - x - p = 0
72. px2 + px - 2 = 0
In Problems 73–78, use the discriminant to determine whether each quadratic equation has two unequal real solutions, a repeated real
solution (a double root), or no real solution, without solving the equation.
73. 2x2 - 6x + 7 = 0
74. x2 + 4x + 7 = 0
75. 9x2 - 30x + 25 = 0
76. 25x2 - 20x + 4 = 0
77. 3x2 + 5x - 8 = 0
78. 2x2 - 3x - 7 = 0
Mixed Practice
In Problems 79–92, find the real solutions, if any, of each equation. Use any method.
79. x2 - 5 = 0
80. x2 - 6 = 0
81. 16x2 - 8x + 1 = 0
82. 9x2 - 12x + 4 = 0
83. 10x2 - 19x - 15 = 0
84. 6x2 + 7x - 20 = 0
85. 2 + z = 6z2
86. 2 = y + 6y2
87. x2 + 22 x =
90. x2 + x = 1
88.
1 2
x = 22 x + 1
2
89. x2 + x = 4
91.
2
7x + 1
x
+
= 2
x - 2
x + 1
x - x - 2
92.
1
2
3x
1
4 - 7x
+
= 2
x + 2
x - 1
x + x - 2
Applications and Extensions
93. Pythagorean Theorem How many right triangles have a
hypotenuse that measures 2x + 3 meters and legs that
measure 2x - 5 meters and x + 7 meters? What are their
dimensions?
94. Pythagorean Theorem How many right triangles have a
hypotenuse that measures 4x + 5 inches and legs that
measure 3x + 13 inches and x inches? What are the
dimensions of the triangle(s)?
95. Dimensions of a Window The area of the opening of a
rectangular window is to be 143 square feet. If the length is
to be 2 feet more than the width, what are the dimensions?
96. Dimensions of a Window The area of a rectangular window
is to be 306 square centimeters. If the length exceeds the
width by 1 centimeter, what are the dimensions?
97. Geometry Find the dimensions of a rectangle whose
perimeter is 26 meters and whose area is 40 square meters.
98. Watering a Field An adjustable water sprinkler that sprays
water in a circular pattern is placed at the center of a square
field whose area is 1250 square feet (see the figure). What is
the shortest radius setting that can be used if the field is to
be completely enclosed within the circle?
99. Constructing a Box An open box is to be constructed from
a square piece of sheet metal by removing a square of side
1 foot from each corner and turning up the edges. If the box
is to hold 4 cubic feet, what should be the dimensions of the
sheet metal?
100. Constructing a Box Rework Problem 99 if the piece of sheet
metal is a rectangle whose length is twice its width.
101. Physics A ball is thrown vertically upward from the
top of a building 96 feet tall with an initial velocity of
SECTION 1.2 Quadratic Equations
80 feet per second. The distance s (in feet) of the ball from
the ground after t seconds is s = 96 + 80t - 16t 2.
(a) After how many seconds does the ball strike the
ground?
(b) After how many seconds will the ball pass the top of the
building on its way down?
102. Physics An object is propelled vertically upward with an
initial velocity of 20 meters per second. The distance s (in
meters) of the object from the ground after t seconds is
s = - 4.9t 2 + 20t.
(a) When will the object be 15 meters above the ground?
(b) When will it strike the ground?
(c) Will the object reach a height of 100 meters?
103. Reducing the Size of a Candy Bar A jumbo chocolate
bar with a rectangular shape measures 12 centimeters
in length, 7 centimeters in width, and 3 centimeters in
thickness. Due to escalating costs of cocoa, management
decides to reduce the volume of the bar by 10%. To
accomplish this reduction, management decides that the
new bar should have the same 3-centimeter thickness, but
the length and width of each should be reduced an equal
number of centimeters. What should be the dimensions of
the new candy bar?
104. Reducing the Size of a Candy Bar Rework Problem 103 if
the reduction is to be 20%.
105. Constructing a Border around a Pool A circular pool
measures 10 feet across. One cubic yard of concrete is to be
used to create a circular border of uniform width around
the pool. If the border is to have a depth of 3 inches, how
wide will the border be? (1 cubic yard = 27 cubic feet) See
the illustration.
x
10 ft
103
109. Comparing Tablets The screen size of a tablet is determined
by the length of the diagonal of the rectangular screen. The
9.7-inch iPad AirTM comes in a 4:3 format, which means
that the ratio of the length to the width of the rectangular
screen is 4:3. What is the area of the iPad’s screen? What
is the area of a 10-inch Google NexusTM if its screen is in a
16:10 format? Which screen is larger? (Hint: If x is the length
3
of a 4:3 format screen, then x is the width.)
4
iPad mini 4:3
Google Nexus 16:10
110. Comparing Tablets Refer to Problem 109. Find the screen
area of a 7.9-inch iPad mini with RetinaTM in a 4:3 format,
and compare it with an 8-inch Dell Venue ProTM if its screen
is in a 16:9 format. Which screen is larger?
111. Field Design A football field is sloped from the center
toward the sides for drainage. The height h, in feet, of the field,
x feet from the side, is given by h = - 0.00025x2 + 0.04x.
Find the height of the field a distance of 35 feet from the
side. Round to the nearest tenth of a foot.
112. College Value The difference, d, in median earnings, in
$1000s, between high school graduates and college graduates
can be approximated by d = - 0.002x2 + 0.319x + 7.512,
where x is the number of years after 1965. Based on this model,
estimate to the nearest year when the difference in median
earnings was $15,000. (Source: Current Population Survey)
113. Student Working A study found that a student’s GPA, g,
is related to the number of hours worked each week, h, by
the equation g = - 0.0006h2 + 0.015h + 3.04. Estimate the
number of hours worked each week for a student with a
GPA of 2.97. Round to the nearest whole hour.
106. Constructing a Border around a Pool Rework Problem 105
if the depth of the border is 4 inches.
114. Fraternity Purchase A fraternity wants to buy a new LED
Smart TV that costs $1470. If 7 members of the fraternity are
not able to contribute, the share for the remaining members
increases by $5. How many members are in the fraternity?
107. Constructing a Border around a Garden A landscaper,
who just completed a rectangular flower garden measuring
6 feet by 10 feet, orders 1 cubic yard of premixed cement,
all of which is to be used to create a border of uniform
width around the garden. If the border is to have a depth
of 3 inches, how wide will the border be? (1 cubic yard =
27 cubic feet)
115. The sum of the consecutive integers 1, 2, 3, c, n is given
1
by the formula n1n + 12. How many consecutive integers,
2
starting with 1, must be added to get a sum of 703?
1
116. Geometry If a polygon of n sides has n1n - 32 diagonals,
2
how many sides will a polygon with 65 diagonals have? Is
there a polygon with 80 diagonals?
10 ft
6 ft
108. Dimensions of a Patio A contractor orders 8 cubic yards
of premixed cement, all of which is to be used to pour a
patio that will be 4 inches thick. If the length of the patio
is specified to be twice the width, what will be the patio
dimensions? (1 cubic yard = 27 cubic feet)
117. Show that the sum of the roots of a quadratic equation
b
is - .
a
118. Show that the product of the roots of a quadratic equation
c
is .
a
119. Find k such that the equation kx2 + x + k = 0 has a repeated
real solution.
120. Find k such that the equation x2 - kx + 4 = 0 has a repeated
real solution.
104
CHAPTER 1 Equations and Inequalities
121. Show that the real solutions of the equation ax2 + bx + c = 0
are the negatives of the real solutions of the equation
ax2 - bx + c = 0. Assume that b2 - 4ac Ú 0.
122. Show that the real solutions of the equation ax2 + bx + c = 0
are the reciprocals of the real solutions of the equation
cx2 + bx + a = 0. Assume that b2 - 4ac Ú 0.
Explaining Concepts: Discussion and Writing
123. Which of the following pairs of equations are equivalent?
Explain.
(a) x2 = 9; x = 3
(b) x = 29; x = 3
(c) 1x - 12 1x - 22 = 1x - 12 2; x - 2 = x - 1
124. Describe three ways that you might solve a quadratic equation.
State your preferred method; explain why you chose it.
125. Explain the benefits of evaluating the discriminant of a
quadratic equation before attempting to solve it.
126. Create three quadratic equations: one having two distinct
real solutions, one having no real solution, and one having
exactly one real solution.
127. The word quadratic seems to imply four (quad), yet a
quadratic equation is an equation that involves a polynomial
of degree 2. Investigate the origin of the term quadratic as
it is used in the expression quadratic equation. Write a brief
essay on your findings.
‘Are You Prepared?’ Answers
1. 1x - 62 1x + 12
2. 12x - 32 1x + 12
5
3. e - , 3 f
3
4. True
5. x2 + 5x +
5 2
25
= ax + b
4
2
1.3 Complex Numbers; Quadratic Equations
in the Complex Number System *
PREPARING FOR THIS SECTION Before getting started, review the following:
r Classification of Numbers (Section R.1, pp. 4–5)
r Rationalizing Denominators (Section R.8, p. 75)
Now Work the ‘Are You Prepared?’ problems on page 111.
OBJECTIVES 1 Add, Subtract, Multiply, and Divide Complex Numbers (p. 105)
2 Solve Quadratic Equations in the Complex Number System (p. 109)
Complex Numbers
One property of a real number is that its square is nonnegative. For example, there
is no real number x for which
x2 = - 1
To remedy this situation, we introduce a new number called the imaginary unit.
DEFINITION
The imaginary unit, which we denote by i, is the number whose square is - 1.
That is,
i2 = - 1
This should not surprise you. If our universe were to consist only of integers,
there would be no number x for which 2x = 1. This was remedied by introducing
1
2
numbers such as and , the rational numbers. If our universe were to consist only
2
3
of rational numbers, there would be no x whose square equals 2. That is, there would
be no number x for which x2 = 2. To remedy this, we introduced numbers such as
3
12 and 2
5, the irrational numbers. Recall that the real numbers consist of the
rational numbers and the irrational numbers. Now, if our universe were to consist
only of real numbers, then there would be no number x whose square is - 1. To
remedy this, we introduce the number i, whose square is - 1.
*This section may be omitted without any loss of continuity.
SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System
105
In the progression outlined, each time we encountered a situation that was
unsuitable, a new number system was introduced to remedy the situation. The number
system that results from introducing the number i is called the complex number system.
DEFINITION
Complex numbers are numbers of the form a + bi, where a and b are real
numbers. The real number a is called the real part of the number a + bi; the
real number b is called the imaginary part of a + bi; and i is the imaginary
unit, so i 2 = - 1.
For example, the complex number - 5 + 6i has the real part - 5 and the
imaginary part 6.
When a complex number is written in the form a + bi, where a and b are real
numbers, it is in standard form. However, if the imaginary part of a complex number
is negative, such as in the complex number 3 + 1 - 22i, we agree to write it instead
in the form 3 - 2i.
Also, the complex number a + 0i is usually written merely as a. This serves to
remind us that the real numbers are a subset of the complex numbers. Similarly, the
complex number 0 + bi is usually written as bi. Sometimes the complex number bi
is called a pure imaginary number.
1 Add, Subtract, Multiply, and Divide Complex Numbers
Equality, addition, subtraction, and multiplication of complex numbers are defined so
as to preserve the familiar rules of algebra for real numbers. Two complex numbers
are equal if and only if their real parts are equal and their imaginary parts are equal.
Equality of Complex Numbers
a + bi = c + di
if and only if
a = c and b = d
(1)
Two complex numbers are added by forming the complex number whose real
part is the sum of the real parts and whose imaginary part is the sum of the imaginary
parts.
Sum of Complex Numbers
1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i
(2)
To subtract two complex numbers, use this rule:
Difference of Complex Numbers
1a + bi2 - 1c + di2 = 1a - c2 + 1b - d2i
EXAMPL E 1
(3)
Adding and Subtracting Complex Numbers
(a) 13 + 5i2 + 1 - 2 + 3i2 = 3 3 + 1 - 22 4 + 15 + 32i = 1 + 8i
(b) 16 + 4i2 - 13 + 6i2 = 16 - 32 + 14 - 62i = 3 + 1 - 22i = 3 - 2i
Now Work
PROBLEM
15
r
106
CHAPTER 1 Equations and Inequalities
Products of complex numbers are calculated as illustrated in Example 2.
EX AMPLE 2
Multiplying Complex Numbers
15 + 3i2 # 12 + 7i2 = 5 # 12 + 7i2 + 3i12 + 7i2 = 10 + 35i + 6i + 21i 2
c
c
Distributive Property
Distributive Property
= 10 + 41i + 211 - 12
c
i2 = - 1
= - 11 + 41i
r
Based on the procedure of Example 2, the product of two complex numbers is
defined as follows:
Product of Complex Numbers
1a + bi2 # 1c + di2 = 1ac - bd2 + 1ad + bc2i
(4)
Do not bother to memorize formula (4). Instead, whenever it is necessary to
multiply two complex numbers, follow the usual rules for multiplying two binomials,
as in Example 2, remembering that i 2 = - 1. For example,
12i2 12i2 = 4i 2 = 41 - 12 = - 4
12 + i2 11 - i2 = 2 - 2i + i - i 2 = 3 - i
Now Work
PROBLEM
21
Algebraic properties for addition and multiplication, such as the commutative,
associative, and distributive properties, hold for complex numbers. The property
that every nonzero complex number has a multiplicative inverse, or reciprocal,
requires a closer look.
DEFINITION
If z = a + bi is a complex number, then its conjugate, denoted by z, is defined as
NOTE The conjugate of a complex number
can be found by changing the sign of the
imaginary part.
■
z = a + bi = a - bi
For example, 2 + 3i = 2 - 3i and - 6 - 2i = - 6 + 2i.
EX A MPLE 3
Multiplying a Complex Number by Its Conjugate
Find the product of the complex number z = 3 + 4i and its conjugate z.
Solution
Since z = 3 - 4i, we have
zz = 13 + 4i2 13 - 4i2 = 9 - 12i + 12i - 16i 2 = 9 + 16 = 25
r
The result obtained in Example 3 has an important generalization.
THEOREM
The product of a complex number and its conjugate is a nonnegative real number.
That is, if z = a + bi, then
zz = a2 + b2
(5)
SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System
107
Proof If z = a + bi, then
zz = 1a + bi2 1a - bi2 = a2 - 1bi2 2 = a2 - b2 i 2 = a2 + b2
■
To express the reciprocal of a nonzero complex number z in standard form,
1
multiply the numerator and denominator of by z. That is, if z = a + bi is a nonzero
z
complex number, then
1
1 z
z
a - bi
1
= = # =
= 2
z
z z
a + bi
zz
a + b2
c
Use (5).
=
EXAMPL E 4
Writing the Reciprocal of a Complex Number in Standard Form
Write
Solution
a
b
- 2
i
2
a + b
a + b2
2
1
in standard form a + bi; that is, find the reciprocal of 3 + 4i.
3 + 4i
The idea is to multiply the numerator and denominator by the conjugate of 3 + 4i,
that is, by the complex number 3 - 4i. The result is
1
1 # 3 - 4i
3 - 4i
3
4
=
=
=
i
3 + 4i
3 + 4i 3 - 4i
9 + 16
25
25
r
To express the quotient of two complex numbers in standard form, multiply the
numerator and denominator of the quotient by the conjugate of the denominator.
EXAMPL E 5
Writing the Quotient of Two Complex Numbers in Standard Form
Write each of the following in standard form.
(a)
Solution
(a)
(b)
1 + 4i
5 - 12i
2 - 3i
4 - 3i
1 + 4i # 5 + 12i
5 + 12i + 20i + 48i 2
1 + 4i
=
=
5 - 12i
5 - 12i 5 + 12i
25 + 144
- 43 + 32i
43
32
=
= +
i
169
169
169
2 - 3i
2 - 3i # 4 + 3i
8 + 6i - 12i - 9i 2
17 - 6i
17
6
=
=
=
=
i
4 - 3i
4 - 3i 4 + 3i
16 + 9
25
25
25
Now Work
EXAMPL E 6
(b)
PROBLEM
r
29
Writing Other Expressions in Standard Form
If z = 2 - 3i and w = 5 + 2i, write each of the following expressions in standard
form.
(a)
z
w
(b) z + w
(c) z + z
108
CHAPTER 1 Equations and Inequalities
Solution
(a)
12 - 3i2 15 - 2i2
z
z#w
10 - 4i - 15i + 6i 2
=
= # =
w
w w
15 + 2i2 15 - 2i2
25 + 4
=
4
19
4 - 19i
=
i
29
29
29
(b) z + w = 12 - 3i2 + 15 + 2i2 = 7 - i = 7 + i
(c) z + z = 12 - 3i2 + 12 + 3i2 = 4
r
The conjugate of a complex number has certain general properties that will be
useful later.
For a real number a = a + 0i, the conjugate is a = a + 0i = a - 0i = a.
THEOREM
The conjugate of a real number is the real number itself.
Other properties that are direct consequences of the definition of the conjugate
are given next. In each statement, z and w represent complex numbers.
THEOREM
The conjugate of the conjugate of a complex number is the complex number
itself.
1z2 = z
(6)
The conjugate of the sum of two complex numbers equals the sum of their
conjugates.
z + w = z + w
(7)
The conjugate of the product of two complex numbers equals the product of
their conjugates.
z#w = z#w
(8)
The proofs of equations (6), (7), and (8) are left as exercises.
Powers of i
The powers of i follow a pattern that is useful to know.
i1
i2
i3
i4
=
=
=
=
i
-1
i2 # i = - 1 # i = - i
i 2 # i 2 = 1 - 12 1 - 12 = 1
i5
i6
i7
i8
=
=
=
=
i4 # i = 1 # i = i
i4 # i2 = - 1
i4 # i3 = - i
i4 # i4 = 1
And so on. The powers of i repeat with every fourth power.
EX AMPLE 7
Evaluating Powers of i
(a) i 27 = i 24 # i 3 = 1i 4 2
(b) i 101 = i 100 # i 1 =
6
# i3 = 16 # i3 = - i
25
1i 4 2 # i = 125 # i = i
r
SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System
EXAMPL E 8
Solution
109
Writing the Power of a Complex Number in Standard Form
Write 12 + i2 3 in standard form.
Use the special product formula for 1x + a2 3.
1x + a2 3 = x3 + 3ax2 + 3a2 x + a3
NOTE Another way to find (2 + i)3 is to
multiply out (2 + i)2 (2 + i).
■
Using this special product formula,
12 + i2 3 = 23 + 3 # i # 22 + 3 # i 2 # 2 + i 3
= 8 + 12i + 61 - 12 + 1 - i2
r
= 2 + 11i
Now Work
PROBLEM
43
2 Solve Quadratic Equations in the Complex Number System
Quadratic equations with a negative discriminant have no real number solution.
However, if we extend our number system to allow complex numbers, quadratic
equations will always have a solution. Since the solution to a quadratic equation
involves the square root of the discriminant, we begin with a discussion of square
roots of negative numbers.
DEFINITION
WARNING In writing 1- N = 1N i,
be sure to place i outside the 1
symbol.
■
EXAMPL E 9
If N is a positive real number, we define the principal square root of −N,
denoted by 1- N , as
2 - N = 2N i
where i is the imaginary unit and i 2 = - 1.
Evaluating the Square Root of a Negative Number
(a) 1- 1 = 11 i = i
(b) 1- 4 = 14 i = 2i
(c) 1- 8 = 18 i = 212 i
EX AM PL E 1 0
r
Solving Equations
Solve each equation in the complex number system.
(a) x2 = 4
Solution
(b) x2 = - 9
(a) x2 = 4
x = { 24 = {2
The equation has two solutions, - 2 and 2. The solution set is 5 - 2, 26 .
(b) x2 = - 9
x = { 2 - 9 = { 29 i = {3i
The equation has two solutions, - 3i and 3i. The solution set is 5 - 3i, 3i6 .
Now Work
PROBLEMS
51
AND
55
r
110
CHAPTER 1 Equations and Inequalities
WARNING When working with square roots of negative numbers, do not set the square
root of a product equal to the product of the square roots (which can be done with positive
real numbers). To see why, look at this calculation: We know that 1100 = 10. However, it is also true
that 100 = 1 - 252 1 - 42, so
10 = 2100 = 21 - 252 1 - 42 = 2- 25 2- 4 =
1 225 i 2 1 24 i 2
= 15i2 12i2 = 10i 2 = - 10
c
Here is the error.
■
Because we have defined the square root of a negative number, we can now
restate the quadratic formula without restriction.
THEOREM
Quadratic Formula
In the complex number system, the solutions of the quadratic equation
ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0, are given
by the formula
x =
EX A MPL E 11
- b { 2b2 - 4ac
2a
(9)
Solving a Quadratic Equation in the Complex Number System
Solve the equation x2 - 4x + 8 = 0 in the complex number system.
Solution
Here a = 1, b = - 4, c = 8, and b2 - 4ac = 1 - 42 2 - 4112 182 = - 16. Using
equation (9), we find that
x =
- 1 - 42 { 2 - 16
2(2 { 2i)
4 { 216 i
4 { 4i
=
=
=
= 2 { 2i
2112
2
2
2
The equation has two solutions: 2 - 2i and 2 + 2i.
The solution set is 5 2 - 2i, 2 + 2i6 .
Check: 2 + 2i:
2 - 2i:
Now Work
12 + 2i2 2 - 412 + 2i2 + 8 = 4 + 8i + 4i 2 - 8 - 8i + 8
=
=
12 - 2i2 2 - 412 - 2i2 + 8 =
=
PROBLEM
4
4
4
4
+
-
4i 2
4 = 0
8i + 4i 2 - 8 + 8i + 8
4 = 0
r
61
The discriminant b2 - 4ac of a quadratic equation still serves as a way to
determine the character of the solutions.
Character of the Solutions of a Quadratic Equation
In the complex number system, consider a quadratic equation ax2 + bx + c = 0
with real coefficients.
1. If b2 - 4ac 7 0, the equation has two unequal real solutions.
2. If b2 - 4ac = 0, the equation has a repeated real solution, a double root.
3. If b2 - 4ac 6 0, the equation has two complex solutions that are not real.
The solutions are conjugates of each other.
SECTION 1.3 Complex Numbers; Quadratic Equations in the Complex Number System
111
The third conclusion in the display is a consequence of the fact that if
b2 - 4ac = - N 6 0, then by the quadratic formula, the solutions are
x =
- b + 2b2 - 4ac
- b + 2- N
- b + 2N i
-b
2N
=
=
=
+
i
2a
2a
2a
2a
2a
and
- b - 2b2 - 4ac
- b - 2- N
- b - 2N i
-b
2N
=
=
=
i
2a
2a
2a
2a
2a
which are conjugates of each other.
x =
Determining the Character of the Solutions of a Quadratic Equation
EX AM PL E 1 2
Without solving, determine the character of the solutions of each equation.
(a) 3x2 + 4x + 5 = 0
Solution
(b) 2x2 + 4x + 1 = 0
(c) 9x2 - 6x + 1 = 0
(a) Here a = 3, b = 4, and c = 5, so b2 - 4ac = 16 - 4132 152 = - 44. The
solutions are two complex numbers that are not real and are conjugates of each
other.
(b) Here a = 2, b = 4, and c = 1, so b2 - 4ac = 16 - 8 = 8. The solutions are
two unequal real numbers.
(c) Here a = 9, b = - 6, and c = 1, so b2 - 4ac = 36 - 4192 112 = 0. The
solution is a repeated real number—that is, a double root.
r
Now Work
PROBLEM
75
1.3 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Name the integers and the rational numbers in the set
6
b - 3, 0, 12, , p r. (pp. 4–5)
5
2. True or False Rational numbers and irrational numbers are
in the set of real numbers. (pp. 4–5)
3
3. Rationalize the denominator of
. (p. 75)
2 + 13
Concepts and Vocabulary
4. In the complex number 5 + 2i, the number 5 is called the
part; the number 2 is called the
part;
the number i is called the
.
5. True or False The conjugate of 2 + 5i is - 2 - 5i.
6. True or False All real numbers are complex numbers.
7. True or False If 2 - 3i is a solution of a quadratic equation
with real coefficients, then - 2 + 3i is also a solution.
8. Which of the following is the principal square root of - 4?
(a) - 2i
(b) 2i
(c) - 2
(d) 2
9. Which operation involving complex numbers requires the
use of a conjugate?
(a) division
(b) multiplication
(c) subtraction
(d) addition
10. Powers of i repeat every
(a) second
(b) third
power.
(c) fourth
(d) fifth
Skill Building
In Problems 11–48, perform the indicated operation, and write each expression in the standard form a + bi.
11. 12 - 3i2 + 16 + 8i2
12. 14 + 5i2 + 1 - 8 + 2i2
13. 1 - 3 + 2i2 - 14 - 4i2
14. 13 - 4i2 - 1 - 3 - 4i2
15. 12 - 5i2 - 18 + 6i2
16. 1 - 8 + 4i2 - 12 - 2i2
17. 312 - 6i2
18. - 412 + 8i2
19. 2i12 - 3i2
20. 3i1 - 3 + 4i2
21. 13 - 4i2 12 + i2
22. 15 + 3i2 12 - i2
23. 1 - 6 + i2 1 - 6 - i2
24. 1 - 3 + i2 13 + i2
25.
10
3 - 4i
26.
13
5 - 12i
29.
6 - i
1 + i
30.
2 + 3i
1 - i
27.
2 + i
i
31. a
1
23 2
+
ib
2
2
28.
2 - i
- 2i
32. a
23
1 2
- ib
2
2
33. 11 + i2 2
34. 11 - i2 2
112
CHAPTER 1 Equations and Inequalities
35. i 23
36. i 14
37. i -15
38. i -23
40. 4 + i 3
41. 6i 3 - 4i 5
42. 4i 3 - 2i 2 + 1
43. 11 + i2 3
45. i 7 11 + i 2 2
46. 2i 4 11 + i 2 2
47. i 6 + i 4 + i 2 + 1
48. i 7 + i 5 + i 3 + i
39. i 6 - 5
44. 13i2 4 + 1
In Problems 49–54, perform the indicated operations, and express your answer in the form a + bi.
49. 2- 4
50. 2- 9
51. 2- 25
52. 2- 64
53. 213 + 4i2 14i - 32
54. 214 + 3i2 13i - 42
In Problems 55–74, solve each equation in the complex number system.
55. x2 + 4 = 0
56. x2 - 4 = 0
57. x2 - 16 = 0
58. x2 + 25 = 0
59. x2 - 6x + 13 = 0
60. x2 + 4x + 8 = 0
61. x2 - 6x + 10 = 0
62. x2 - 2x + 5 = 0
63. 8x2 - 4x + 1 = 0
64. 10x2 + 6x + 1 = 0
65. 5x2 + 1 = 2x
66. 13x2 + 1 = 6x
67. x2 + x + 1 = 0
68. x2 - x + 1 = 0
69. x3 - 8 = 0
70. x3 + 27 = 0
71. x4 = 16
72. x4 = 1
73. x4 + 13x2 + 36 = 0
74. x4 + 3x2 - 4 = 0
In Problems 75–80, without solving, determine the character of the solutions of each equation in the complex number system.
75. 3x2 - 3x + 4 = 0
76. 2x2 - 4x + 1 = 0
77. 2x2 + 3x = 4
78. x2 + 6 = 2x
79. 9x2 - 12x + 4 = 0
80. 4x2 + 12x + 9 = 0
81. 2 + 3i is a solution of a quadratic equation with real
coefficients. Find the other solution.
82. 4 - i is a solution of a quadratic equation with real
coefficients. Find the other solution.
In Problems 83–86, z = 3 - 4i and w = 8 + 3i. Write each expression in the standard form a + bi.
83. z + z
85. zz
84. w - w
86. z - w
Applications and Extensions
87. Electrical Circuits The impedance Z, in ohms, of a circuit
element is defined as the ratio of the phasor voltage V, in
volts, across the element to the phasor current I, in amperes,
V
through the elements. That is, Z = . If the voltage across a
I
circuit element is 18 + i volts and the current through the
element is 3 - 4 i amperes, determine the impedance.
88. Parallel Circuits In an ac circuit with two parallel pathways,
the total impedance Z, in ohms, satisfies the formula
1
1
1
=
+
, where Z1 is the impedance of the first pathway
Z
Z1
Z2
and Z2 is the impedance of the second pathway. Determine
the total impedance if the impedances of the two pathways
are Z1 = 2 + i ohms and Z2 = 4 - 3i ohms.
89. Use z = a + bi to show that z + z = 2a and z - z = 2bi.
90. Use z = a + bi to show that z = z.
91. Use z = a + bi and w = c + di to show
that z + w = z + w.
92. Use z = a + bi and w = c + di to show that z # w = z # w.
Explaining Concepts: Discussion and Writing
93. Explain to a friend how you would add two complex
numbers and how you would multiply two complex numbers.
Explain any differences between the two explanations.
94. Write a brief paragraph that compares the method used to
rationalize the denominator of a radical expression and the
method used to write the quotient of two complex numbers
in standard form.
96. Explain how the method of multiplying two complex
numbers is related to multiplying two binomials.
97. What Went Wrong? A student multiplied 2- 9 and 2- 9
as follows:
2- 9 # 2- 9 = 2( - 9)( - 9)
= 281
95. Use an Internet search engine to investigate the origins of
complex numbers. Write a paragraph describing what you
find, and present it to the class.
= 9
The instructor marked the problem incorrect. Why?
‘Are You Prepared?’ Answers
6
1. Integers: 5 - 3, 06; rational numbers: b - 3, 0, r
5
2. True
3. 3 1 2 - 23 2
SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations
113
1.4 Radical Equations; Equations Quadratic in Form;
Factorable Equations
PREPARING FOR THIS SECTION Before getting started, review the following:
r Square Roots (Section R.2, pp. 23–24)
r Factoring Polynomials (Section R.5, pp. 49–55)
Now Work the ‘Are You Prepared?’ problems on page 117.
r nth Roots; Rational Exponents (Section R.8,
pp. 73–77)
OBJECTIVES 1 Solve Radical Equations (p. 113)
2 Solve Equations Quadratic in Form (p. 114)
3 Solve Equations by Factoring (p. 116)
1 Solve Radical Equations
When the variable in an equation occurs in a square root, cube root, and so on—that
is, when it occurs in a radical—the equation is called a radical equation. Sometimes
a suitable operation will change a radical equation to one that is linear or quadratic.
A commonly used procedure is to isolate the most complicated radical on one side
of the equation and then eliminate it by raising each side to a power equal to the
index of the radical. Care must be taken, however, because apparent solutions
that are not, in fact, solutions of the original equation may result. These are called
extraneous solutions. Therefore, we need to check all answers when working with
radical equations, and we check them in the original equation.
EXAMPL E 1
Solving a Radical Equation
3
Find the real solutions of the equation: 2
2x - 4 - 2 = 0
Solution
The equation contains a radical whose index is 3. Isolate it on the left side.
3
2
2x - 4 - 2 = 0
3
2
2x - 4 = 2
Add 2 to both sides.
Now raise each side to the third power (the index of the radical is 3) and solve.
3
12
2x
- 42
2x - 4
2x
x
3
=
=
=
=
23
8
12
6
Raise each side to the power 3.
Simplify.
Add 4 to both sides.
Divide both sides by 2.
3
3
3
Check: 2
2162 - 4 - 2 = 2
12 - 4 - 2 = 2
8 - 2 = 2 - 2 = 0
The solution set is 5 66 .
Now Work
EXAMPL E 2
r
PROBLEM
9
Solving a Radical Equation
Find the real solutions of the equation: 2x - 1 = x - 7
Solution
Square both sides since the index of a square root is 2.
2x - 1
- 122
x - 1
2
x - 15x + 50
1 2x
=
=
=
=
x - 7
1x - 72 2
x2 - 14x + 49
0
Square both sides.
Remove parentheses.
Put in standard form.
114
CHAPTER 1 Equations and Inequalities
1x - 102 1x - 52 = 0
x = 10 or x = 5
Factor.
Apply the Zero-Product Property and solve.
Check:
x = 10:
2x - 1 = 210 - 1 = 29 = 3 and x - 7 = 10 - 7 = 3
x = 5:
2x - 1 = 25 - 1 = 24 = 2 and x - 7 = 5 - 7 = - 2
The solution x = 5 does not check, so it is extraneous; the only solution of the
equation is x = 10. The solution set is {10}.
r
Now Work
PROBLEM
19
Sometimes it is necessary to raise each side to a power more than once in order
to solve a radical equation.
Solving a Radical Equation
EX AMPLE 3
Find the real solutions of the equation: 22x + 3 - 2x + 2 = 2
First, isolate the more complicated radical expression (in this case, 12x + 3) on the
left side.
Solution
22x + 3 = 2x + 2 + 2
Now square both sides (the index of the radical on the left is 2).
1 22x
+ 32 =
2
2x + 3 =
1 2x
1 2x
+ 2 + 222
+ 2 2 + 42x + 2 + 4
2
Square both sides.
Multiply out.
2x + 3 = x + 2 + 42x + 2 + 4
Simplify.
2x + 3 = x + 6 + 42x + 2
Combine like terms.
Because the equation still contains a radical, isolate the remaining radical on the
right side and again square both sides.
x - 3 = 42x + 2
1x - 32 2 = 16(x + 2)
2
x - 6x + 9
x - 22x - 23
1x - 232 1x + 12
x = 23 or x
2
=
=
=
=
16x + 32
0
0
-1
Isolate the radical on the right side.
Square both sides.
Multiply out.
Put in standard form.
Factor.
Apply the Zero-Product Property and solve.
The original equation appears to have the solution set 5 - 1, 236 . However, we have
not yet checked.
Check:
x = 23:
22x + 3 - 2x + 2 = 221232 + 3 - 223 + 2 = 249 - 225 = 7 - 5 = 2
x = - 1:
22x + 3 - 2x + 2 = 221 - 12 + 3 - 2 - 1 + 2 = 21 - 21 = 1 - 1 = 0
The equation has only one solution, 23; the solution - 1 is extraneous. The solution
set is {23}.
r
Now Work
PROBLEM
31
2 Solve Equations Quadratic in Form
The equation x4 + x2 - 12 = 0 is not quadratic in x, but it is quadratic in x2. That
is, if we let u = x2, we get u2 + u - 12 = 0, a quadratic equation. This equation
can be solved for u, and in turn, by using u = x2, we can find the solutions x of the
original equation.
SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations
115
In general, if an appropriate substitution u transforms an equation into one of
the form
au2 + bu + c = 0
a ≠ 0
then the original equation is called an equation of the quadratic type or an equation
quadratic in form.
The difficulty of solving such an equation lies in the determination that the
equation is, in fact, quadratic in form. After you are told an equation is quadratic in
form, it is easy enough to see it, but some practice is needed to enable you to recognize
such equations on your own.
EXAMPL E 4
Solution
Solving an Equation Quadratic in Form
Find the real solutions of the equation: 1x + 22 2 + 111x + 22 - 12 = 0
For this equation, let u = x + 2. Then u2 = 1x + 22 2, and the original equation,
1x + 22 2 + 111x + 22 - 12 = 0
becomes
u2 + 11u - 12 = 0
1u + 122 1u - 12 = 0
u = - 12 or u = 1
Let u = x + 2. Then u 2 = (x + 2)2.
Factor.
Solve.
But we want to solve for x. Because u = x + 2, we have
x + 2 = - 12 or x + 2 = 1
x = - 14
x = -1
Check: x = - 14:
x = - 1:
1 - 14 + 22 2 + 111 - 14 + 22 - 12
= 1 - 122 2 + 111 - 122 - 12 = 144 - 132 - 12 = 0
1 - 1 + 22 2 + 111 - 1 + 22 - 12 = 1 + 11 - 12 = 0
The original equation has the solution set 5 - 14, - 16 .
EXAMPL E 5
r
Solving an Equation Quadratic in Form
Find the real solutions of the equation: 1x2 - 12 + 1x2 - 12 - 12 = 0
2
Solution
For the equation 1x2 - 12 + 1x2 - 12 - 12 = 0,
2
that u2 = 1x2 - 12 . Then the original equation,
2
let
u = x2 - 1
so
1x2 - 12 + 1x2 - 12 - 12 = 0
2
becomes
u2 + u - 12 = 0
1u + 42 1u - 32 = 0
u = - 4 or u = 3
2
Let u = x 2 - 1. Then u 2 = (x 2 - 12 .
Factor.
Solve.
But remember that we want to solve for x. Because u = x2 - 1, we have
x2 - 1 = - 4 or x2 - 1 = 3
x2 = - 3
x2 = 4
The first of these has no real solution; the second has the solution set 5 - 2, 26 .
Check: x = - 2:
x = 2:
1( - 2)2 - 12 2 + 1( - 2)2 - 12 - 12 = 9 + 3 - 12 = 0
1(2)2 - 12 2 + 1(2)2 - 12 - 12 = 9 + 3 - 12 = 0
The original equation has the solution set { - 2, 2}.
r
116
CHAPTER 1 Equations and Inequalities
EX AMPLE 6
Solving an Equation Quadratic in Form
Find the real solutions of the equation: x + 21x - 3 = 0
Solution
For the equation x + 21x - 3 = 0, let u = 1x. Then u2 = x, and the original
equation,
x + 21x - 3 = 0
becomes
u2 + 2u - 3 = 0
Let u = 1x. Then u 2 = x.
1u + 32 1u - 12 = 0
Factor.
u = - 3 or u = 1
Solve.
Since u = 1x, we have 1x = - 3 or 1x = 1. The first of these, 1x = - 3, has no
real solution, since the principal square root of a real number is never negative. The
second, 1x = 1, has the solution x = 1.
Check: 1 + 221 - 3 = 1 + 2 - 3 = 0
r
The original equation has the solution set {1}.
A N OT H E R M E T H O D F O R S O LV I N G E XA M P L E
6
WOULD
B E TO T R E AT I T A S A R A D I C A L E Q UAT I O N . S O LV E I T T H I S
WAY F O R P R A C T I C E .
The idea should now be clear. If an equation contains an expression and that
same expression squared, make a substitution for the expression. You may get a
quadratic equation.
Now Work
PROBLEM
53
3 Solve Equations by Factoring
We have already solved certain quadratic equations using factoring. Let’s look at
examples of other kinds of equations that can be solved by factoring.
EX AMPLE 7
Solving an Equation by Factoring
Solve the equation: x4 = 4x2
Solution
Begin by collecting all terms on one side. This results in 0 on one side and an
expression to be factored on the other.
x4 = 4x2
x4 - 4x2 = 0
x2 1x2 - 42 = 0
Factor.
x = 0
or x - 4 = 0
Apply the Zero-Product Property.
x = 0
x = 4
2
x = 0
or
x = 2:
2
or
x = - 2 or x = 2
Check: x = - 2:
x = 0:
2
1 - 22 4 = 16 and 41 - 22 2 = 16
4
0
4
2
The solution set is 5 - 2, 0, 26 .
= 0 and 4 # 02 = 0
= 16 and 4 # 22 = 16
- 2 is a solution.
0 is a solution.
2 is a solution.
r
SECTION 1.4 Radical Equations; Equations Quadratic in Form; Factorable Equations
EXAMPL E 8
117
Solving an Equation by Factoring
Solve the equation: x3 - x2 - 4x + 4 = 0
Solution
Do you recall the method of factoring by grouping? (If not, review pp. 53–54.)
Group the terms of x3 - x2 - 4x + 4 = 0 as follows:
1x3 - x2 2 - 14x - 42 = 0
Factor out x2 from the first grouping and 4 from the second.
x2 1x - 12 - 41x - 12 = 0
This reveals the common factor 1x - 12, so we have
1x2
1x - 22 1x
x - 2 =
x =
- 42 1x - 12
+ 22 1x - 12
0 or x + 2
2
x
=
=
=
=
0
0
Factor again.
0 or x - 1 = 0 Set each factor equal to 0.
-2
x = 1 Solve.
Check:
x = - 2:
1 - 22 3 - 1 - 22 2 - 41 - 22 + 4 = - 8 - 4 + 8 + 4 = 0
- 2 is a solution.
x = 1:
1 - 1 - 4112 + 4 = 1 - 1 - 4 + 4 = 0
1 is a solution.
x = 2:
2 - 2 - 4122 + 4 = 8 - 4 - 8 + 4 = 0
2 is a solution.
3
3
2
2
The solution set is 5 - 2, 1, 26 .
Now Work
PROBLEM
r
81
1.4 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. True or False The principal square root of any nonnegative
real number is always nonnegative. (pp. 23–24)
3
2. 2
-8 =
(pp. 73–77)
3. Factor 6x3 - 2x2 (pp. 49–55)
Concepts and Vocabulary
4. True or False Factoring can only be used to solve quadratic
equations or equations that are quadratic in form.
5. If u is an expression that involves x, then the equation
au2 + bu + c = 0, a ≠ 0, is called an equation
.
6. True or False Radical equations sometimes have extraneous
solutions.
7. An apparent solution that does not satisfy the original
equation is called a(n)
solution.
(a) extraneous
(b) imaginary
(c) radical
(d) conditional
8. Which equation is likely to require squaring each side more
than once?
(a) 2x + 2 = 23x - 5
(b) x4 - 3x2 = 10
(c) 2x + 1 + 2x - 4 = 8
(d) 23x + 1 = 5
Skill Building
In Problems 9–42, find the real solutions of each equation.
9. 22t - 1 = 1
10. 23t + 4 = 2
11. 23t + 4 = - 6
12. 25t + 3 = - 2
3
13. 2
1 - 2x - 3 = 0
3
14. 2
1 - 2x - 1 = 0
5 2
15. 2
x + 2x = - 1
4 2
16. 2
x + 16 = 25
17. x = 82x
18. x = 31x
19. 215 - 2x = x
20. 212 - x = x
21. x = 22x - 1
22. x = 22- x - 1
23. 2x2 - x - 4 = x + 2
24. 23 - x + x2 = x - 2
25. 3 + 23x + 1 = x
26. 2 + 212 - 2x = x
27. 23(x + 10) - 4 = x
28. 21 - x - 3 = x + 2
29. 22x + 3 - 2x + 1 = 1
30. 23x + 7 + 2x + 2 = 1
31. 23x + 1 - 2x - 1 = 2
32. 23x - 5 - 2x + 7 = 2
118
CHAPTER 1 Equations and Inequalities
33. 23 - 21x = 1x
37. 15x - 22 1>3 = 2
36. 13x - 52 1>2 = 2
39. 1x2 + 92
1>2
= 5
35. 13x + 12 1>2 = 4
34. 210 + 31x = 1x
40. 1x2 - 162
1>2
38. 12x + 12 1>3 = - 1
41. x3>2 - 3x1>2 = 0
= 9
42. x3>4 - 9x1>4 = 0
In Problems 43–74, find the real solutions of each equation.
43. x4 - 5x2 + 4 = 0
44. x4 - 10x2 + 25 = 0
45. 3x4 - 2x2 - 1 = 0
46. 2x4 - 5x2 - 12 = 0
47. x6 + 7x3 - 8 = 0
48. x6 - 7x3 - 8 = 0
49. 1x + 22 + 71x + 22 + 12 = 0
50. 12x + 52 - 12x + 52 - 6 = 0
51. 13x + 42 2 - 613x + 42 + 9 = 0
52. 12 - x2 2 + 12 - x2 - 20 = 0
53. 21s + 12 2 - 51s + 12 = 3
54. 311 - y2 2 + 511 - y2 + 2 = 0
55. x - 4x1x = 0
56. x + 81x = 0
57. x + 1x = 20
2
2
58. x + 1x = 6
59. t
61. 4x1>2 - 9x1>4 + 4 = 0
62. x1>2 - 3x1>4 + 2 = 0
4
63. 25x2 - 6 = x
4
64. 2
4 - 5x2 = x
65. x2 + 3x + 2x2 + 3x = 6
66. x2 - 3x - 2x2 - 3x = 2
67.
1
1
=
+ 2
x + 1
1x + 12 2
68.
- 2t
1>4
+ 1 = 0
60. z1>2 - 4z1>4 + 4 = 0
1
1
+
= 12
x - 1
1x - 12 2
69. 3x -2 - 7x -1 - 6 = 0
72. 3x4>3 + 5x2>3 - 2 = 0
71. 2x2>3 - 5x1>3 - 3 = 0
70. 2x -2 - 3x -1 - 4 = 0
73. a
1>2
2
2v
v
b +
= 8
v + 1
v + 1
74. a
2
y
y
b = 6a
b + 7
y - 1
y - 1
In Problems 75–90, find the real solutions of each equation by factoring.
75. x3 - 9x = 0
77. 4x3 = 3x2
76. x4 - x2 = 0
78. x5 = 4x3
79. x3 + x2 - 20x = 0
80. x3 + 6x2 - 7x = 0
81. x3 + x2 - x - 1 = 0
82. x3 + 4x2 - x - 4 = 0
83. x3 - 3x2 - 4x + 12 = 0
84. x3 - 3x2 - x + 3 = 0
85. 2x3 + 4 = x2 + 8x
86. 3x3 + 4x2 = 27x + 36
87. 5x3 + 45x = 2x2 + 18
88. 3x3 + 12x = 5x2 + 20
89. x1x2 - 3x2 1>3 + 21x2 - 3x2 4>3 = 0
90. 3x1x2 + 2x2 1>2 - 21x2 + 2x2 3>2 = 0
In Problems 91–96, find the real solutions of each equation. Use a calculator to express any solutions rounded to two decimal places.
91. x - 4x1>2 + 2 = 0
92. x2>3 + 4x1>3 + 2 = 0
93. x4 + 23 x2 - 3 = 0
94. x4 + 22 x2 - 2 = 0
95. p11 + t2 2 = p + 1 + t
96. p11 + r2 2 = 2 + p11 + r2
Mixed Practice
97. If k =
x + 3
and k 2 - k = 12, find x.
x - 3
98. If k =
x + 3
and k 2 - 3k = 28, find x.
x - 4
Applications
99. Physics: Using Sound to Measure Distance The distance to the surface of
the water in a well can sometimes be found by dropping an object into the
well and measuring the time elapsed until a sound is heard. If t 1 is the time
(measured in seconds) that it takes for the object to strike the water, then t 1
will obey the equation s = 16t 21 , where s is the distance (measured in feet). It
1s
follows that t 1 =
. Suppose that t 2 is the time that it takes for the sound
4
of the impact to reach your ears. Because sound waves are known to travel
at a speed of approximately 1100 feet per second, the time t 2 to travel the
s
. See the illustration.
distance s will be t 2 =
1100
Sound waves:
s
t2 ––––
1100
Falling object:
s
t1 ––
4
SECTION 1.5 Solving Inequalities
119
Now t 1 + t 2 is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have
the equation
Total time elapsed =
s
1s
+
4
1100
Find the distance to the water’s surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.
100. Crushing Load A civil engineer relates the thickness T, in inches, and height H, in feet, of a square wooden pillar to its crushing
2
4 LH
. If a square wooden pillar is 4 inches thick and 10 feet high, what is its crushing load?
load L, in tons, using the model T =
A 25
101. Foucault’s Pendulum The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The
l
period T, in seconds, is given by the formula T = 2p
, where l is the length, in feet, of the pendulum. In 1851, Jean Bernard
A 32
Leon Foucault demonstrated the axial rotation of Earth using a large pendulum that he hung in the Panthéon in Paris. The period
of Foucault’s pendulum was approximately 16.5 seconds. What was its length?
Explaining Concepts: Discussion and Writing
102. Make up a radical equation that has no solution.
103. Make up a radical equation that has an extraneous solution.
104. Discuss the step in the solving process for radical equations
that leads to the possibility of extraneous solutions. Why is
there no such possibility for linear and quadratic equations?
105. What Went Wrong? On an exam, Jane solved the
equation 22x + 3 - x = 0 and wrote that the solution set
was { - 1, 3}. Jane received 3 out of 5 points for the problem.
Jane asks you why she received 3 out of 5 points. Provide an
explanation.
‘Are You Prepared?’ Answers
1. True
2. - 2
3. 2x2 13x - 12
1.5 Solving Inequalities
PREPARING FOR THIS SECTION Before getting started, review the following:
r Algebra Essentials (Section R.2, pp. 17–26)
Now Work the ‘Are You Prepared?’ problems on page 127.
OBJECTIVES 1
2
3
4
Use Interval Notation (p. 120)
Use Properties of Inequalities (p. 121)
Solve Inequalities (p. 123)
Solve Combined Inequalities (p. 124)
Suppose that a and b are two real numbers and a 6 b. The notation a 6 x 6 b
means that x is a number between a and b. The expression a 6 x 6 b is equivalent
to the two inequalities a 6 x and x 6 b. Similarly, the expression a … x … b is
equivalent to the two inequalities a … x and x … b. The remaining two possibilities,
a … x 6 b and a 6 x … b, are defined similarly.
Although it is acceptable to write 3 Ú x Ú 2, it is preferable to reverse the
inequality symbols and write instead 2 … x … 3 so that the values go from smaller
to larger, reading from left to right.
A statement such as 2 … x … 1 is false because there is no number x for which
2 … x and x … 1. Finally, never mix inequality symbols, as in 2 … x Ú 3.
120
CHAPTER 1 Equations and Inequalities
1 Use Interval Notation
Let a and b represent two real numbers with a 6 b.
DEFINITION
In Words
The notation [a, b] represents all
real numbers between a and b,
inclusive. The notation (a, b)
represents all real numbers between
a and b, not including either a or b.
An open interval, denoted by (a, b), consists of all real numbers x for which
a 6 x 6 b.
A closed interval, denoted by [a, b], consists of all real numbers x for which
a … x … b.
The half-open, or half-closed, intervals are (a, b], consisting of all real
numbers x for which a 6 x … b, and [a, b), consisting of all real numbers x
for which a … x 6 b.
In each of these definitions, a is called the left endpoint and b the right endpoint of
the interval.
The symbol q (read as “infinity”) is not a real number, but notation used to
indicate unboundedness in the positive direction. The symbol - q (read as “negative
infinity”) also is not a real number, but notation used to indicate unboundedness in
the negative direction. The symbols q and - q are used to define five other kinds
of intervals:
[a, ˆ )
(a, ˆ )
(− ˆ , a]
(− ˆ , a)
(− ˆ , ˆ )
Consists of all real numbers x for which x
Consists of all real numbers x for which x
Consists of all real numbers x for which x
Consists of all real numbers x for which x
Consists of all real numbers
Ú
7
…
6
a
a
a
a
Note that q and - q are never included as endpoints, since neither is a real number.
Table 1 summarizes interval notation, corresponding inequality notation, and
their graphs.
Table 1
EX AMPLE 1
Interval
Inequality
Graph
The open interval (a, b)
a<x<b
The closed interval [a, b]
a#x#b
The half-open interval [a, b)
a#x<b
The half-open interval (a, b]
a<x#b
The interval [a, ∞)
x$a
The interval (a, ∞)
x>a
The interval (−∞, a]
x#a
The interval (−∞, a)
x<a
The interval (−∞, ∞)
All real numbers
a
b
a
b
a
b
a
b
a
a
a
a
Writing Inequalities Using Interval Notation
Write each inequality using interval notation.
(a) 1 … x … 3
Solution
(b) - 4 6 x 6 0
(c) x 7 5
(d) x … 1
(a) 1 … x … 3 describes all real numbers x between 1 and 3, inclusive. In interval
notation, we write 3 1, 34 .
(b) In interval notation, - 4 6 x 6 0 is written 1 - 4, 02.
SECTION 1.5 Solving Inequalities
(c) In interval notation, x 7 5 is written 15, q 2.
(d) In interval notation, x … 1 is written 1 - q , 14 .
EXAMPL E 2
r
Writing Intervals Using Inequality Notation
Write each interval as an inequality involving x.
(a) 3 1, 42
Solution
121
(a)
(b)
(c)
(d)
(b) 12, q 2
(c) 3 2, 34
(d) 1 - q , - 34
3 1, 42 consists of all real numbers x for which 1 … x 6 4.
12, q 2 consists of all real numbers x for which x 7 2.
3 2, 34 consists of all real numbers x for which 2 … x … 3.
1 - q , - 34 consists of all real numbers x for which x … - 3.
Now Work
PROBLEMS
13, 25,
AND
r
33
2 Use Properties of Inequalities
The product of two positive real numbers is positive, the product of two negative
real numbers is positive, and the product of 0 and 0 is 0. For any real number a, the
value of a2 is 0 or positive; that is, a2 is nonnegative. This is called the nonnegative
property.
Nonnegative Property
In Words
The square of a real number is
never negative.
For any real number a,
a2 Ú 0
(1)
When the same number is added to both sides of an inequality, an equivalent
inequality is obtained. For example, since 3 6 5, then 3 + 4 6 5 + 4 or 7 6 9. This
is called the addition property of inequalities.
In Words
The addition property states
that the sense, or direction, of
an inequality remains unchanged
if the same number is added to
each side.
Addition Property of Inequalities
For real numbers a, b, and c,
If a 6 b, then a + c 6 b + c.
If a 7 b, then a + c 7 b + c.
(2a)
(2b)
Figure 2 illustrates the addition property (2a). In Figure 2(a), we see that a lies
to the left of b. If c is positive, then a + c and b + c lie c units to the right of a and c
units to the right of b, respectively. Consequently, a + c must lie to the left of b + c;
that is, a + c 6 b + c. Figure 2(b) illustrates the situation if c is negative.
– c units
c units
c units
a
b
a +c
– c units
b +c
(a) If a < b and c > 0,
then a + c < b + c.
a+c
b +c a
b
(b) If a < b and c < 0,
then a + c < b + c.
Figure 2 Addition property of inequalities
D R AW A N I L LU S T R AT I O N S I M I L A R TO F I G U R E
T H AT I L LU S T R AT E S T H E A D D I T I O N P R O P E R T Y
2
(2b).
122
CHAPTER 1 Equations and Inequalities
EX AMPLE 3
Addition Property of Inequalities
(a) If x 6 - 5, then x + 5 6 - 5 + 5 or x + 5 6 0.
(b) If x 7 2, then x + 1 - 22 7 2 + 1 - 22 or x - 2 7 0.
Now Work
EX AMPLE 4
PROBLEM
r
41
Multiplying an Inequality by a Positive Number
Express as an inequality the result of multiplying each side of the inequality 3 6 7
by 2.
Solution
Begin with
3 6 7
Multiplying each side by 2 yields the numbers 6 and 14, so we have
r
6 6 14
EX AMPLE 5
Multiplying an Inequality by a Negative Number
Express as an inequality the result of multiplying each side of the inequality 9 7 2
by - 4.
Solution
Begin with
9 7 2
Multiplying each side by - 4 yields the numbers - 36 and - 8, so we have
r
- 36 6 - 8
In Words
Multiplying by a negative number
reverses the inequality.
Note that the effect of multiplying both sides of 9 7 2 by the negative
number - 4 is that the direction of the inequality symbol is reversed.
Examples 4 and 5 illustrate the following general multiplication properties for
inequalities:
Multiplication Properties for Inequalities
In Words
The multiplication properties
state that the sense, or direction,
of an inequality remains the same if
each side is multiplied by a positive
real number, whereas the direction
is reversed if each side is multiplied
by a negative real number.
EX A MPLE 6
For real numbers a, b, and c,
If a 6 b and if c 7 0, then ac 6 bc.
If a 6 b and if c 6 0, then ac 7 bc.
(3a)
If a 7 b and if c 7 0, then ac 7 bc.
If a 7 b and if c 6 0, then ac 6 bc.
(3b)
Multiplication Property of Inequalities
1
1
12x2 6 162 or x 6 3.
2
2
x
x
(b) If
7 12, then - 3a
b 6 - 31122 or x 6 - 36.
-3
-3
- 4x
-8
(c) If - 4x 6 - 8, then
7
or x 7 2.
-4
-4
(d) If - x 7 8, then 1 - 12 1 - x2 6 1 - 12 182 or x 6 - 8.
(a) If 2x 6 6, then
Now Work
PROBLEM
47
r
SECTION 1.5 Solving Inequalities
In Words
The reciprocal property states
that the reciprocal of a positive
real number is positive and that
the reciprocal of a negative real
number is negative.
123
Reciprocal Property for Inequalities
1
7 0.
a
1
If a 6 0, then 6 0.
a
If a 7 0, then
1
7 0, then a 7 0.
a
1
If 6 0, then a 6 0.
a
If
(4a)
(4b)
3 Solve Inequalities
An inequality in one variable is a statement involving two expressions, at least
one containing the variable, separated by one of the inequality symbols: 6 , … , 7 ,
or Ú . To solve an inequality means to find all values of the variable for which the
statement is true. These values are called solutions of the inequality.
For example, the following are all inequalities involving one variable x:
x + 5 6 8
2x - 3 Ú 4
x2 - 1 … 3
x + 1
7 0
x - 2
As with equations, one method for solving an inequality is to replace it by a
series of equivalent inequalities until an inequality with an obvious solution, such as
x 6 3, is obtained. Equivalent inequalities are obtained by applying some of the same
properties that are used to find equivalent equations. The addition property and the
multiplication properties for inequalities form the basis for the following procedures.
Procedures That Leave the Inequality Symbol Unchanged
1. Simplify both sides of the inequality by combining like terms and eliminating
parentheses:
Replace
by
x + 2 + 6 7 2x + 51x + 12
x + 8 7 7x + 5
2. Add or subtract the same expression on both sides of the inequality:
Replace
by
3x - 5 6 4
13x - 52 + 5 6 4 + 5
3. Multiply or divide both sides of the inequality by the same positive
expression:
Replace
4x 7 16 by
4x
16
7
4
4
Procedures That Reverse the Sense or Direction of the Inequality Symbol
1. Interchange the two sides of the inequality:
Replace
3 6 x by x 7 3
2. Multiply or divide both sides of the inequality by the same negative
expression:
Replace
- 2x 7 6 by
- 2x
6
6
-2
-2
As the examples that follow illustrate, we solve inequalities using many of
the same steps that we would use to solve equations. In writing the solution of an
124
CHAPTER 1 Equations and Inequalities
inequality, either set notation or interval notation may be used, whichever is more
convenient.
EX AMPLE 7
Solving an Inequality
Solve the inequality 3 - 2x 6 5, and graph the solution set.
Solution
3 - 2x 6 5
3 - 2x - 3 6 5 - 3
Subtract 3 from both sides.
- 2x 6 2
Simplify.
- 2x
2
7
-2
-2
Divide both sides by - 2. (The sense
of the inequality symbol is reversed.)
x 7 -1
3
2
1
0
1
2
Simplify.
The solution set is 5 x x 7 - 16 or, using interval notation, all numbers in the
interval 1 - 1, q 2. See Figure 3 for the graph.
r
Figure 3 x 7 - 1
EX AMPLE 8
Solving an Inequality
Solve the inequality 4x + 7 Ú 2x - 3, and graph the solution set.
Solution
4x + 7 Ú 2x - 3
4x + 7 - 7 Ú 2x - 3 - 7
Subtract 7 from both sides.
4x Ú 2x - 10
Simplify.
4x - 2x Ú 2x - 10 - 2x
2x Ú - 10
Simplify.
2x
- 10
Ú
2
2
Divide both sides by 2. (The direction
of the inequality symbol is unchanged.)
x Ú -5
6
5
4
3
2
1
Subtract 2x from both sides.
Simplify.
The solution set is 5 x x Ú - 56 or, using interval notation, all numbers in the
interval 3 - 5, q 2. See Figure 4 for the graph.
r
Figure 4 x Ú - 5
Now Work
PROBLEM
55
4 Solve Combined Inequalities
EX AMPLE 9
Solving a Combined Inequality
Solve the inequality - 5 6 3x - 2 6 1, and graph the solution set.
Solution
Recall that the inequality
- 5 6 3x - 2 6 1
is equivalent to the two inequalities
- 5 6 3x - 2 and 3x - 2 6 1
SECTION 1.5 Solving Inequalities
125
Solve each of these inequalities separately.
-5
-5 + 2
-3
-3
3
-1
6 3x - 2
6 3x - 2 + 2
6 3x
3x
6
3
6 x
Add 2 to both sides.
Simplify.
Divide both sides by 3.
Simplify.
3x - 2 6 1
3x - 2 + 2 6 1 + 2
3x 6 3
3x
3
6
3
3
x 6 1
The solution set of the original pair of inequalities consists of all x for which
- 1 6 x and x 6 1
3
2
1
0
1
This may be written more compactly as 5 x - 1 6 x 6 16 . In interval notation, the
solution is 1 - 1, 12. See Figure 5 for the graph.
2
r
Figure 5 - 1 6 x 6 1
Observe in the preceding process that solving each of the two inequalities required
exactly the same steps. A shortcut to solving the original inequality algebraically is to
deal with the two inequalities at the same time, as follows:
-5
-5 + 2
-3
-3
3
-1
6 3x - 2 6 1
6 3x - 2 + 2 6 1 + 2 Add 2 to each part.
6
3x
6 3
Simplify.
3x
3
6
6
Divide each part by 3.
3
3
Simplify.
6
x
6 1
Solving a Combined Inequality
EX AM PL E 1 0
Solve the inequality - 1 …
Solution
-1 …
3 - 5x
2
… 9
3 - 5x
b … 2(9)
2
Multiply each part by 2 to remove the denominator.
3 - 5x
Simplify.
2( - 12 … 2a
-2 …
3 - 5x
… 9, and graph the solution set.
2
… 18
- 2 - 3 … 3 - 5x - 3 … 18 - 3 Subtract 3 from each part to isolate
4
3
2
1
Figure 6 - 3 … x … 1
0
1
2
-5 …
- 5x
… 15
-5
Ú
-5
- 5x
-5
Ú
the term containing x.
Simplify.
15
-5
Divide each part by - 5 (reverse the direction
of each inequality symbol).
1 Ú
x
Ú -3
Simplify.
-3 …
x
… 1
Reverse the order so that the numbers
get larger as you read from left to right.
The solution set is 5 x - 3 … x … 16 , that is, all x in the interval 3 - 3, 14 . Figure 6
illustrates the graph.
r
Now Work
PROBLEM
75
126
CHAPTER 1 Equations and Inequalities
EX A MPL E 11
Solution
Using the Reciprocal Property to Solve an Inequality
Solve the inequality 14x - 12 -1 7 0, and graph the solution set.
1
Recall that 14x - 12 -1 =
. The Reciprocal Property states that when
4x
- 1
1
7 0, then a 7 0.
a
14x - 12 -1 7 0
1
7 0
4x - 1
4x - 1 7 0
Reciprocal Property
4x 7 1
Add 1 to both sides.
x 7
0
1
–
4
1
1
4
Divide both sides by 4.
1
1
The solution set is e x 0 x 7 f, that is, all x in the interval a , q b . Figure 7
4
4
illustrates the graph.
r
1
Figure 7 x 7
4
Now Work
EX A MPL E 1 2
PROBLEM
85
Creating Equivalent Inequalities
If - 1 6 x 6 4, find a and b so that a 6 2x + 1 6 b.
Solution
The idea here is to change the middle part of the combined inequality from x to
2x + 1, using properties of inequalities.
-1 6
x
6 4
- 2 6 2x 6 8
- 1 6 2x + 1 6 9
Multiply each part by 2.
Add 1 to each part.
r
Now we see that a = - 1 and b = 9.
Now Work
PROBLEM
95
Application
EX A MPL E 1 3
Physics: Ohm’s Law
In electricity, Ohm’s law states that E = IR, where E is the voltage (in volts), I is the
current (in amperes), and R is the resistance (in ohms). An air-conditioning unit is
rated at a resistance of 10 ohms. If the voltage varies from 110 to 120 volts, inclusive,
what corresponding range of current will the air conditioner draw?
Solution
The voltage lies between 110 and 120, inclusive, so
110
110
110
110
10
11
… E
… IR
… I 1102
I 1102
…
10
…
I
… 120
… 120
… 120
120
…
10
… 12
Ohm’s law, E = IR
R = 10
Divide each part by 10.
Simplify.
The air conditioner will draw between 11 and 12 amperes of current, inclusive.
r
SECTION 1.5 Solving Inequalities
127
1.5 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Graph the inequality: x Ú - 2. (pp. 17–26)
2. True or False - 5 7 - 3 (pp. 17–26)
Concepts and Vocabulary
3. A(n)
, denoted 3a, b4, consists of all
real numbers x for which a … x … b.
9. True or False The square of any real number is always
nonnegative.
10. True or False A half-closed interval must have an endpoint
of either - q or q .
state that the sense,
4. The
or direction, of an inequality remains the same if each side
is multiplied by a positive number, while the direction is
reversed if each side is multiplied by a negative number.
11. Which of the following will change the direction, or sense, of
an inequality?
(a) Dividing each side by a positive number
(b) Interchanging sides
(c) Adding a negative number to each side
(d) Subtracting a positive number from each side
In Problems 5–8, assume that a 6 b and c 6 0.
5. True or False a + c 6 b + c
6. True or False a - c 6 b - c
7. True or False ac 7 bc
12. Which pair of inequalities is equivalent to 0 6 x … 3?
(a) x 7 0 and x Ú 3
(b) x 6 0 and x Ú 3
(c) x 7 0 and x … 3
(d) x 6 0 and x … 3
b
a
6
8. True or False
c
c
Skill Building
In Problems 13–18, express the graph shown in blue using interval notation. Also express each as an inequality involving x.
13.
14.
–1
0
1
2
3
–2
–1
0
1
2
16.
15.
–2
–1
0
1
2
–1
0
1
2
3
17.
–1
0
1
2
3
–1
0
1
2
3
18.
In Problems 19–24, an inequality is given. Write the inequality obtained by:
(a) Adding 3 to each side of the given inequality.
(b) Subtracting 5 from each side of the given inequality.
(c) Multiplying each side of the given inequality by 3.
(d) Multiplying each side of the given inequality by - 2.
19. 3 6 5
21. 4 7 - 3
20. 2 7 1
22. - 3 7 - 5
23. 2x + 1 6 2
24. 1 - 2x 7 5
In Problems 25–32, write each inequality using interval notation, and graph each inequality on the real number line.
25. 0 … x … 4
26. - 1 6 x 6 5
27. 4 … x 6 6
28. - 2 6 x 6 0
29. x Ú 4
30. x … 5
31. x 6 - 4
32. x 7 1
In Problems 33–40, write each interval as an inequality involving x, and graph each inequality on the real number line.
33. 32, 54
34. 11, 22
35. 1 - 3, - 22
36. 30, 12
37. 34, q 2
38. 1 - q , 24
39. 1 - q , - 32
40. 1 - 8, q 2
In Problems 41–54, fill in the blank with the correct inequality symbol.
41. If x 6 5, then x - 5
42. If x 6 - 4, then x + 4
0.
43. If x 7 - 4, then x + 4
0.
0.
44. If x 7 6, then x - 6
0.
45. If x Ú - 4, then 3x
- 12.
46. If x … 3, then 2x
47. If x 7 6, then - 2x
- 12.
48. If x 7 - 2, then - 4x
8.
49. If x Ú 5, then - 4x
- 20.
50. If x … - 4, then - 3x
12.
51. If 2x 7 6, then x
53. If -
1
x … 3, then x
2
52. If 3x … 12, then x
3.
- 6.
54. If -
1
x 7 1, then x
4
6.
4.
- 4.
128
CHAPTER 1 Equations and Inequalities
In Problems 55–92, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.
55. x + 1 6 5
56. x - 6 6 1
57. 1 - 2x … 3
58. 2 - 3x … 5
59. 3x - 7 7 2
60. 2x + 5 7 1
61. 3x - 1 Ú 3 + x
62. 2x - 2 Ú 3 + x
63. - 21x + 32 6 8
64. - 311 - x2 6 12
65. 4 - 311 - x2 … 3
66. 8 - 412 - x2 … - 2x
67.
1
1x - 42 7 x + 8
2
68. 3x + 4 7
70.
x
x
Ú 2 +
3
6
71. 0 … 2x - 6 … 4
72. 4 … 2x + 2 … 10
74. - 3 … 3 - 2x … 9
75. - 3 6
73. - 5 … 4 - 3x … 2
76. 0 6
3x + 2
6 4
2
77. 1 6 1 -
1
1x - 22
3
69.
1
x 6 4
2
80. 1x - 12 1x + 12 7 1x - 32 1x + 42
82. x19x - 52 … 13x - 12 2
83.
85. 14x + 22 -1 6 0
86. 12x - 12 -1 7 0
88. 2(3x + 5)-1 … - 3
89. 0 6
1
x + 1
3
…
6
2
3
4
1
x 6 1
3
81. x14x + 32 … 12x + 12 2
84.
1
x + 1
2
6
…
3
2
3
87. (2 - 7x)-1 Ú 5
3
2
6
x
5
90. 0 6
92. 0 6 13x + 62 -1 6
1
2
2x - 1
6 0
4
78. 0 6 1 -
79. 1x + 22 1x - 32 7 1x - 12 1x + 12
91. 0 6 12x - 42 -1 6
x
x
Ú 1 2
4
4
2
6
x
3
1
3
Applications and Extensions
In Problems 93–102, find a and b.
93. If - 1 6 x 6 1, then a 6 x + 4 6 b.
94. If - 3 6 x 6 2, then a 6 x - 6 6 b.
95. If 2 6 x 6 3, then a 6 - 4x 6 b.
1
96. If - 4 6 x 6 0, then a 6 x 6 b.
2
97. If 0 6 x 6 4, then a 6 2x + 3 6 b.
98. If - 3 6 x 6 3, then a 6 1 - 2x 6 b.
99. If - 3 6 x 6 0, then a 6
100. If 2 6 x 6 4, then a 6
1
6 b.
x + 4
30-year-old female in 2014 could expect to live at least 55.6
more years.
(a) To what age could an average 30-year-old male expect
to live? Express your answer as an inequality.
(b) To what age could an average 30-year-old female expect
to live? Express your answer as an inequality.
(c) Who can expect to live longer, a male or a female? By
how many years?
Source: Social Security Administration, 2014
1
6 b.
x - 6
101. If 6 6 3x 6 12, then a 6 x2 6 b.
102. If 0 6 2x 6 6, then a 6 x2 6 b.
103. What is the domain of the variable in the expression
13x + 6 ?
104. What is the domain of the variable in the expression
18 + 2x ?
105. A young adult may be defined as someone older than 21
but less than 30 years of age. Express this statement using
inequalities.
106. Middle-aged may be defined as being 40 or more and less
than 60. Express this statement using inequalities.
107. Life Expectancy The Social Security Administration
determined that an average 30-year-old male in 2014 could
expect to live at least 51.9 more years and that an average
JAN
2014
JULY
2069
NOV
2065
108. General Chemistry For a certain ideal gas, the volume V
(in cubic centimeters) equals 20 times the temperature T (in
degrees Celsius). If the temperature varies from 80° to
120°C, inclusive, what is the corresponding range of the
volume of the gas?
109. Real Estate A real estate agent agrees to sell an apartment
complex according to the following commission schedule:
$45,000 plus 25% of the selling price in excess of $900,000.
Assuming that the complex will sell at some price between
SECTION 1.5 Solving Inequalities
$900,000 and $1,100,000, inclusive, over what range does the
agent’s commission vary? How does the commission vary as
a percent of selling price?
110. Sales Commission A used car salesperson is paid a commission
of $25 plus 40% of the selling price in excess of owner’s cost.
The owner claims that used cars typically sell for at least
owner’s cost plus $200 and at most owner’s cost plus $3000.
For each sale made, over what range can the salesperson
expect the commission to vary?
111. Federal Tax Withholding The percentage method of
withholding for federal income tax (2014) states that a single
person whose weekly wages, after subtracting withholding
allowances, are over $753, but not over $1762, shall have
$97.75 plus 25% of the excess over $753 withheld. Over what
range does the amount withheld vary if the weekly wages
vary from $900 to $1100, inclusive?
Source: Employer’s Tax Guide. Internal Revenue Service, 2014.
112. Exercising Sue wants to lose weight. For healthy weight
loss, the American College of Sports Medicine (ACSM)
recommends 200 to 300 minutes of exercise per week. For
the first six days of the week, Sue exercised 40, 45, 0, 50,
25, and 35 minutes. How long should Sue exercise on the
seventh day in order to stay within the ACSM guidelines?
113. Electricity Rates Commonwealth Edison Company’s charge
for electricity in January 2014 was 8.21¢ per kilowatt-hour.
In addition, each monthly bill contains a customer charge of
$15.37. If last year’s bills ranged from a low of $72.84 to a high
of $237.04, over what range did usage vary (in kilowatt-hours)?
Source: Commonwealth Edison Co., 2014.
114. Water Bills The Village of Oak Lawn charges homeowners
$57.07 per quarter-year plus $5.81 per 1000 gallons for water
usage in excess of 10,000 gallons. In 2014 one homeowner’s
quarterly bill ranged from a high of $150.03 to a low of
$97.74. Over what range did water usage vary?
Source: Village of Oak Lawn, Illinois, January 2014.
115. Markup of a New Car The markup over dealer’s cost of
a new car ranges from 12% to 18%. If the sticker price is
$18,000, over what range will the dealer’s cost vary?
116. IQ Tests A standard intelligence test has an average score
of 100. According to statistical theory, of the people who take
the test, the 2.5% with the highest scores will have scores of
more than 1.96s above the average, where s (sigma, a number
called the standard deviation) depends on the nature of the
test. If s = 12 for this test and there is (in principle) no upper limit to the score possible on the test, write the interval of
possible test scores of the people in the top 2.5%.
129
117. Computing Grades In your Economics 101 class, you have
scores of 68, 82, 87, and 89 on the first four of five tests. To get
a grade of B, the average of the first five test scores must be
greater than or equal to 80 and less than 90.
(a) Solve an inequality to find the least score you can get on
the last test and still earn a B.
(b) What score do you need if the fifth test counts double?
What do I need to get a B?
82
68 89
87
118. “Light” Foods For food products to be labeled “light,” the
U.S. Food and Drug Administration requires that the altered
product must either contain at least one-third fewer calories
than the regular product or contain at least one-half less fat
than the regular product. If a serving of Miracle Whip® Light
contains 20 calories and 1.5 grams of fat, then what must be
true about either the number of calories or the grams of fat
in a serving of regular Miracle Whip®?
a + b
6 b. The
119. Arithmetic Mean If a 6 b, show that a 6
2
a+b
number
is called the arithmetic mean of a and b.
2
120. Refer to Problem 119. Show that the arithmetic mean of a
and b is equidistant from a and b.
121. Geometric Mean If 0 6 a 6 b, show that a 6 1ab 6 b.
The number 1ab is called the geometric mean of a and b.
122. Refer to Problems 119 and 121. Show that the geometric
mean of a and b is less than the arithmetic mean of a and b.
123. Harmonic Mean For 0 6 a 6 b, let h be defined by
1 1
1
1
= a + b
h
2 a
b
Show that a 6 h 6 b. The number h is called the harmonic
mean of a and b.
124. Refer to Problems 119, 121, and 123. Show that the harmonic
mean of a and b equals the geometric mean squared, divided
by the arithmetic mean.
125. Another Reciprocal Property Prove that if 0 6 a 6 b, then
1
1
6 .
0 6
b
a
Explaining Concepts: Discussion and Writing
126. Make up an inequality that has no solution. Make up an
inequality that has exactly one solution.
127. The inequality x2 + 1 6 - 5 has no real solution. Explain why.
128. Do you prefer to use inequality notation or interval notation
to express the solution to an inequality? Give your reasons.
Are there particular circumstances when you prefer one to
the other? Cite examples.
‘Are You Prepared?’ Answers
1.
4
2
2. False
0
129. How would you explain to a fellow student the underlying
reason for the multiplication properties for inequalities
(page 122)? That is, the sense or direction of an inequality
remains the same if each side is multiplied by a positive real
number, whereas the direction is reversed if each side is
multiplied by a negative real number.
130
CHAPTER 1 Equations and Inequalities
1.6 Equations and Inequalities Involving Absolute Value
PREPARING FOR THIS SECTION Before getting started, review the following:
r Algebra Essentials (Chapter R, Section R.2, pp. 17–26)
Now Work the ‘Are You Prepared?’ problems on page 132.
OBJECTIVES 1 Solve Equations Involving Absolute Value (p. 130)
2 Solve Inequalities Involving Absolute Value (p. 130)
1 Solve Equations Involving Absolute Value
Recall that on the real number line, the absolute value of a equals the distance from
the origin to the point whose coordinate is a. For example, there are two points
whose distance from the origin is 5 units, - 5 and 5. So the equation 0 x 0 = 5 will
have the solution set 5 - 5, 56 . This leads to the following result:
THEOREM
If a is a positive real number and if u is any algebraic expression, then
0 u 0 = a is equivalent to u = a or u = - a
(1)
Equation (1) requires that a be a positive number. If a = 0, equation (1) becomes
|u| = 0 which is equivalent to u = 0. If a is less than zero, the equation has no
real solution. To see why, consider the equation |x| = - 2. Because there is no real
number whose distance from 0 on the real number line is negative, this equation has
no real solution.
EX AMPLE 1
Solving an Equation Involving Absolute Value
Solve the equations:
(a) 0 x + 4 0 = 13
Solution
(b) 0 2x - 3 0 + 2 = 7
(a) This follows the form of equation (1), where u = x + 4. There are two possibilities:
x + 4 = 13 or x + 4 = - 13
x = 9 or
x = - 17
The solution set is 5 - 17, 96 .
(b) The equation 0 2x - 3 0 + 2 = 7 is not in the form of equation (1). Proceed as
follows:
0 2x - 3 0 + 2 = 7
0 2x - 3 0 = 5
2x - 3 = 5
2x = 8
x = 4
or
or
or
Subtract 2 from each side.
2x - 3 = - 5 Apply (1).
2x = - 2 Add 3 to both sides.
x = - 1 Divide both sides by 2.
The solution set is 5 - 1, 46 .
Now Work
PROBLEM
r
11
2 Solve Inequalities Involving Absolute Value
EX AMPLE 2
Solving an Inequality Involving Absolute Value
Solve the inequality: 0 x 0 6 4
SECTION 1.6 Equations and Inequalities Involving Absolute Value
Solution
Less than 4 units
from origin O
0
1
2
We are looking for all points whose coordinate x is a distance less than 4 units from
the origin. See Figure 8 for an illustration. Because any number x between - 4 and
4 satisfies the condition 0 x 0 6 4, the solution set consists of all numbers x for which
- 4 6 x 6 4, that is, all x in the interval 1 - 4, 42.
r
O
25 24 23 22 21
131
3
4
Figure 8 x 6 4
THEOREM If a is a positive number and if u is an algebraic expression, then
0 u 0 6 a is equivalent to - a 6 u 6 a
0 u 0 … a is equivalent to - a … u … a
(2)
(3)
In other words, 0 u 0 6 a is equivalent to - a 6 u and u 6 a.
–a
0
a
Figure 9 u … a, a 7 0
See Figure 9 for an illustration of statement (3).
EXAMPL E 3
Solving an Inequality Involving Absolute Value
Solve the inequality 0 2x + 4 0 … 3, and graph the solution set.
0 2x + 4 0 … 3
Solution
-3
-3 - 4
-7
-7
2
7
2
5
7
–2
2
1
–2 0
2
4
EXAMPL E 4
1 3– 2
Figure 11 1 - 4x 6 5
Subtract 4 from each part.
Simplify.
Divide each part by 2.
Simplify.
7
1
7
1
… x … - f, that is, all x in the interval c - , - d . See
2
2
2
2
Figure 10 for the graph of the solution set.
Solving an Inequality Involving Absolute Value
Solve the inequality 0 1 - 4x 0 6 5, and graph the solution set.
0 1 - 4x 0 6 5
-5
-5 - 1
-6
-6
-4
3
2
2
Apply statement (3).
The solution set is e x ` -
Solution
0
… 2x + 4 … 3
… 2x + 4 - 4 … 3 - 4
…
2x
… -1
2x
-1
…
…
2
2
1
x
…
… 2
r
Figure 10 2x + 4 … 3
5 4 3 2 1
This follows the form of statement (3);
the expression u = 2x + 4 is inside the
absolute value bars.
3
4
6 1 - 4x 6 5
6 1 - 4x - 1 6 5 - 1
6
- 4x
6 4
- 4x
4
7
7
-4
-4
7
x
7 -1
-1 6
x
6
3
2
This expression follows the form of statement (2);
the expression u = 1 - 4x is inside the absolute
value bars.
Apply statement (2).
Subtract 1 from each part.
Simplify.
Divide each part by - 4, which reverses the
direction of the inequality symbols.
Simplify.
Rearrange the ordering.
The solution set is e x 0 - 1 6 x 6
3
3
f, that is, all x in the interval a - 1, b . See
2
2
Figure 11 for the graph of the solution set.
r
Now Work
PROBLEM
41
132
CHAPTER 1 Equations and Inequalities
EX AMPLE 5
Solution
5 4 3 2 1
0
1
2
3
4
Solving an Inequality Involving Absolute Value
Solve the inequality 0 x 0 7 3, and graph the solution set.
We are looking for all points whose coordinate x is a distance greater
than 3 units from the origin. Figure 12 illustrates the situation. Any number
x less than - 3 or greater than 3 satisfies the condition 0 x 0 7 3. The
solution set consists of all numbers x for which x 6 - 3 or x 7 3, that is, all x in
1 - q , - 32 h 13, q 2.*
r
Figure 12 x 7 3
THEOREM
If a is a positive number and u is an algebraic expression, then
0 u 0 7 a is equivalent to u 6 - a or u 7 a
(4)
0 u 0 Ú a is equivalent to u … - a or u Ú a
–a
0
a
Figure 13 u Ú a, a 7 0
See Figure 13 for an illustration of statement (5).
Solving an Inequality Involving Absolute Value
EX AMPLE 6
Solve the inequality 0 2x - 5 0 7 3, and graph the solution set.
0 2x - 5 0 7 3
Solution
2x - 5
2x - 5 + 5
2x
2x
2
x
2 1
0
1
2
(5)
3
4
5
6
7
Figure 14 2x - 5 7 3
This follows the form of statement (4); the expression
u = 2x - 5 is inside the absolute value bars.
6 -3
or
2x - 5 7
6 - 3 + 5 or 2x - 5 + 5 7
6 2
or
2x 7
2
2x
6
or
7
2
2
6 1
or
x 7
3
Apply statement (4).
3 + 5 Add 5 to each part.
8
Simplify.
8
Divide each part by 2.
2
4
Simplify.
The solution set is 5 x x 6 1 or x 7 46 , that is, all x in 1 - q , 12 h 14, q 2. See
Figure 14 for the graph of the solution set.
r
WARNING A common error to be avoided is to attempt to write the solution x 6 1 or x 7 4
as the combined inequality 1 7 x 7 4, which is incorrect, since there are no numbers x for which
1 7 x and x 7 4. Another common error is to “mix” the symbols and write 1 6 x 7 4, which
makes no sense.
■
Now Work
PROBLEM
45
*Recall that the symbol h stands for the union of two sets. Refer to page 2 if necessary.
1.6 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. 0 - 2 0 =
2. True or False 0 x 0 Ú 0 for any real number x. (p. 19)
(p. 19)
Concepts and Vocabulary
3. The solution set of the equation 0 x 0 = 5 is 5
4. The solution set of the inequality 0 x 0 6 5 is
{x 0
}.
5. True or False The equation 0 x 0 = - 2 has no solution.
6.
6. True or False The inequality 0 x 0 Ú - 2 has the set of real
numbers as its solution set.
7. Which of the following pairs of inequalities is equivalent
to 0 x 0 7 4?
(a) x 7 - 4 and x 6 4
(b) x 6 - 4 and x 6 4
(c) x 7 - 4 or x 7 4
(d) x 6 - 4 or x 7 4
8. Which of the following has no solution?
(a) 0 x 0 6 - 5 (b) 0 x 0 … 0 (c) 0 x 0 7 0
(d) 0 x 0 Ú 0
SECTION 1.6 Equations and Inequalities Involving Absolute Value
133
Skill Building
In Problems 9–36, find the real solutions, if any, of each equation.
9. 0 2x 0 = 6
13. 0 1 - 4t 0 + 8 = 13
10. 0 3x 0 = 12
14. 0 1 - 2z 0 + 6 = 9
11. 0 2x + 3 0 = 5
17. 0 - 2 0 x = 4
18. 0 3 0 x = 9
19.
22. `
23. 0 u - 2 0 = -
21. `
2
x
+ ` = 2
3
5
x
1
- ` = 1
2
3
12. 0 3x - 1 0 = 2
16. 0 - x 0 = 0 1 0
15. 0 - 2x 0 = 0 8 0
2
0x0 = 9
3
20.
3
0x0 = 9
4
24. 0 2 - v 0 = - 1
1
2
25. 4 - 0 2x 0 = 3
1
26. 5 - ` x ` = 3
2
27. 0 x2 - 9 0 = 0
28. 0 x2 - 16 0 = 0
29. 0 x2 - 2x 0 = 3
30. 0 x2 + x 0 = 12
31. 0 x2 + x - 1 0 = 1
32. 0 x2 + 3x - 2 0 = 2
33. `
34. `
35. 0 x2 + 3x 0 = 0 x2 - 2x 0
36. 0 x2 - 2x 0 = 0 x2 + 6x 0
3x - 2
` = 2
2x - 3
2x + 1
` = 1
3x + 4
In Problems 37–64, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.
37. 0 2x 0 6 8
38. 0 3x 0 6 15
39. 0 3x 0 7 12
40. 0 2x 0 7 6
41. 0 x - 2 0 + 2 6 3
42. 0 x + 4 0 + 3 6 5
43. 0 3t - 2 0 … 4
45. 0 2x - 3 0 Ú 2
46. 0 3x + 4 0 Ú 2
47. 0 1 - 4x 0 - 7 6 - 2
44. 0 2u + 5 0 … 7
49. 0 1 - 2x 0 7 3
50. 0 2 - 3x 0 7 1
51. 0 - 4x 0 + 0 - 5 0 … 1
52. 0 - x 0 - 0 4 0 … 2
53. 0 - 2x 0 7 0 - 3 0
54. 0 - x - 2 0 Ú 1
55. - 0 2x - 1 0 Ú - 3
56. - 0 1 - 2x 0 Ú - 3
57. 0 2x 0 6 - 1
58. 0 3x 0 Ú 0
59. 0 5x 0 Ú - 1
60. 0 6x 0 6 - 2
61. `
62. 3 - 0 x + 1 0 6
2x + 3
1
- ` 6 1
3
2
1
2
63. 5 + 0 x - 1 0 7
1
2
48. 0 1 - 2x 0 - 4 6 - 1
64. `
2x - 3
1
+ ` 7 1
2
3
Applications and Extensions
65. Body Temperature “Normal” human body temperature is
98.6°F. If a temperature x that differs from normal by at least
1.5° is considered unhealthy, write the condition for an unhealthy
temperature x as an inequality involving an absolute value, and
solve for x.
66. Household Voltage In the United States, normal household
voltage is 110 volts. However, it is not uncommon for
actual voltage to differ from normal voltage by at most
5 volts. Express this situation as an inequality involving an
absolute value. Use x as the actual voltage and solve for x.
67. Reading Books A HuffPost/YouGov poll conducted
September 27–28, 2013, found that Americans read an
average of 13.6 books per year. Suppose HuffPost/YouGov
is 99% confident that the result from this poll is off by fewer
than 1.8 books from the actual average x. Express this
situation as an inequality involving absolute value, and solve
the inequality for x to determine the interval in which the
actual average is likely to fall. [Note: In statistics, this interval is
called a 99% confidence interval.]
68. Speed of Sound According to data from the Hill Aerospace
Museum (Hill Air Force Base, Utah), the speed of sound
varies depending on altitude, barometric pressure, and
temperature. For example, at 20,000 feet, 13.75 inches of
mercury, and - 12.3°F, the speed of sound is about 707 miles
per hour, but the speed can vary from this result by as much
as 55 miles per hour as conditions change.
(a) Express this situation as an inequality involving an
absolute value.
(b) Using x for the speed of sound, solve for x to find an
interval for the speed of sound.
1
69. Express the fact that x differs from 3 by less than as an
2
inequality involving an absolute value. Solve for x.
70. Express the fact that x differs from - 4 by less than 1 as an
inequality involving an absolute value. Solve for x.
71. Express the fact that x differs from - 3 by more than 2 as an
inequality involving an absolute value. Solve for x.
72. Express the fact that x differs from 2 by more than 3 as an
inequality involving an absolute value. Solve for x.
134
CHAPTER 1 Equations and Inequalities
82. Prove that 0 a - b 0 Ú 0 a 0 - 0 b 0 .
In Problems 73–78, find a and b.
73. If 0 x - 1 0 6 3, then a 6 x + 4 6 b.
[Hint: Apply the triangle inequality from Problem 81 to
0 a 0 = 0 1a - b2 + b 0 .]
74. If 0 x + 2 0 6 5, then a 6 x - 2 6 b.
75. If 0 x + 4 0 … 2, then a … 2x - 3 … b.
83. If a 7 0, show that the solution set of the inequality
76. If 0 x - 3 0 … 1, then a … 3x + 1 … b.
77. If 0 x - 2 0 … 7, then a …
x2 6 a
1
… b.
x - 10
consists of all numbers x for which
- 1a 6 x 6 1a
1
… b.
78. If 0 x + 1 0 … 3, then a …
x + 5
84. If a 7 0, show that the solution set of the inequality
x2 7 a
79. Show that: if a 7 0, b 7 0, and 1a 6 1b, then a 6 b.
[Hint: b - a = 1 1b - 1a2 1 1b + 1a2 .]
consists of all numbers x for which
80. Show that a … 0 a 0 .
x 6 - 1a or x 7 1a
81. Prove the triangle inequality 0 a + b 0 … 0 a 0 + 0 b 0 .
[Hint: Expand 0 a + b 0 2 = 1a + b2 2, and use the result of
Problem 80.]
In Problems 85–92, use the results found in Problems 83 and 84 to solve each inequality.
85. x2 6 1
86. x2 6 4
87. x2 Ú 9
88. x2 Ú 1
89. x2 … 16
90. x2 … 9
91. x2 7 4
92. x2 7 16
93. Solve 0 3x - 0 2x + 1 0 0 = 4.
94. Solve 0 x + 0 3x - 2 0 0 = 2.
Explaining Concepts: Discussion and Writing
95. The equation 0 x 0 = - 2 has no solution. Explain why.
96. The inequality 0 x 0 7 - 0.5 has all real numbers as solutions.
Explain why.
97. The inequality 0 x 0 7 0 has as solution set 5 x x ≠ 06.
Explain why.
‘Are You Prepared?’ Answers
1. 2
2. True
1.7 Problem Solving: Interest, Mixture, Uniform Motion,
Constant Rate Job Applications
OBJECTIVES 1
2
3
4
5
Translate Verbal Descriptions into Mathematical Expressions (p. 135)
Solve Interest Problems (p. 136)
Solve Mixture Problems (p. 137)
Solve Uniform Motion Problems (p. 138)
Solve Constant Rate Job Problems (p. 140)
Applied (word) problems do not come in the form “Solve the equation .c” Instead,
they supply information using words, a verbal description of the real problem. So, to
solve applied problems, we must be able to translate the verbal description into the
language of mathematics. This can be done by using variables to represent unknown
quantities and then finding relationships (such as equations) that involve these
variables. The process of doing all this is called mathematical modeling. An equation
or inequality that describes a relationship among the variables is called a model.
SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
135
Any solution to the mathematical problem must be checked against the
mathematical problem, the verbal description, and the real problem. See Figure 15
for an illustration of the modeling process.
Real
problem
Verbal
description
Language of
mathematics
Mathematical
problem
Check
Check
Check
Solution
Figure 15
1 Translate Verbal Descriptions into Mathematical Expressions
EXAMPL E 1
Translating Verbal Descriptions into Mathematical Expressions
(a) For uniform motion, the average speed of an object equals the distance traveled
divided by the time required.
Translation: If r is the speed, d the distance, and t the time, then r =
d
.
t
(b) Let x denote a number.
The number 5 times as large as x is 5x.
The number 3 less than x is x - 3.
The number that exceeds x by 4 is x + 4.
The number that, when added to x, gives 5 is 5 - x.
Now Work
PROBLEM
r
9
Always check the units used to measure the variables of an applied problem.
In Example (1a), if r is measured in miles per hour, then the distance d must be
expressed in miles, and the time t must be expressed in hours. It is a good practice
to check units to be sure that they are consistent and make sense.
The steps to follow for solving applied problems, given earlier, are repeated
next.
Steps for Solving Applied Problems
STEP 1:
STEP 2:
STEP 3:
STEP 4:
STEP 5:
Read the problem carefully, perhaps two or three times. Pay particular
attention to the question being asked in order to identify what you are
looking for. Identify any relevant formulas you may need (d = rt,
A = pr 2, etc.). If you can, determine realistic possibilities for the answer.
Assign a letter (variable) to represent what you are looking for, and if
necessary, express any remaining unknown quantities in terms of this
variable.
Make a list of all the known facts, and translate them into mathematical
expressions. These may take the form of an equation or an inequality
involving the variable. If possible, draw an appropriately labeled
diagram to assist you. Sometimes, creating a table or chart helps.
Solve for the variable, and then answer the question.
Check the answer with the facts in the problem. If it agrees,
congratulations! If it does not agree, try again.
136
CHAPTER 1 Equations and Inequalities
2 Solve Interest Problems
Interest is money paid for the use of money. The total amount borrowed (whether
by an individual from a bank in the form of a loan, or by a bank from an
individual in the form of a savings account) is called the principal. The rate of
interest, expressed as a percent, is the amount charged for the use of the principal
for a given period of time, usually on a yearly (that is, per annum) basis.
Simple Interest Formula
If a principal of P dollars is borrowed for a period of t years at a per annum
interest rate r, expressed as a decimal, the interest I charged is
I = Prt
(1)
Interest charged according to formula (1) is called simple interest. When using
formula (1), be sure to express r as a decimal. For example, if the rate of interest is
4%, then r = 0.04.
EX AMPLE 2
Finance: Computing Interest on a Loan
Suppose that Juanita borrows $500 for 6 months at the simple interest rate of 9%
per annum. What is the interest that Juanita will be charged on the loan? How much
does Juanita owe after 6 months?
Solution
The rate of interest is given per annum, so the actual time that the money is borrowed
must be expressed in years. The interest charged would be the principal, $500, times
1
the rate of interest (9, = 0.09), times the time in years, :
2
1
Interest charged = I = Prt = 15002 10.092 a b = +22.50
2
After 6 months, Juanita will owe what she borrowed plus the interest:
r
+500 + +22.50 = +522.50
EX AMPLE 3
Financial Planning
Candy has $70,000 to invest and wants an annual return of $2800, which requires an
overall rate of return of 4%. She can invest in a safe, government-insured certificate
of deposit, but it pays only 2%. To obtain 4%, she agrees to invest some of her money
in noninsured corporate bonds paying 7%. How much should be placed in each
investment to achieve her goal?
Solution
STEP 1: The question is asking for two dollar amounts: the principal to invest in the
corporate bonds and the principal to invest in the certificate of deposit.
STEP 2: Let b represent the amount (in dollars) to be invested in the bonds. Then
70,000 - b is the amount that will be invested in the certificate. (Do you
see why?)
STEP 3: We set up a table:
Bonds
Certificate
Total
Principal ($)
Rate
Time (yr)
b
7% = 0.07
1
0.07b
Interest ($)
70,000 - b
2% = 0.02
1
0.02(70,000 - b)
70,000
4% = 0.04
1
0.04(70,000) = 2800
SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
137
Since the combined interest from the investments is equal to the total interest,
we have
Bond interest + Certificate interest = Total interest
0.07b + 0.02 170,000 - b2 = 2800
(Note that the units are consistent: the unit is dollars on each side.)
STEP 4: 0.07b + 1400 - 0.02b = 2800
0.05b = 1400
b = 28,000
Candy should place $28,000 in the bonds and +70,000 - +28,000 = +42,000
in the certificate.
STEP 5: The interest on the bonds after 1 year is 0.07 1+28,0002 = +1960; the
interest on the certificate after 1 year is 0.021+42,0002 = +840. The total
annual interest is $2800, the required amount.
r
Now Work
PROBLEM
19
3 Solve Mixture Problems
Oil refineries sometimes produce gasoline that is a blend of two or more types of
fuel; bakeries occasionally blend two or more types of flour for their bread. These
problems are referred to as mixture problems because they combine two or more
quantities to form a mixture.
EX AMPL E 4
Blending Coffees
The manager of a Starbucks store decides to experiment with a new blend of coffee.
She will mix some B grade Colombian coffee that sells for $5 per pound with some
A grade Arabica coffee that sells for $10 per pound to get 100 pounds of the new
blend. The selling price of the new blend is to be $7 per pound, and there is to be no
difference in revenue between selling the new blend and selling the other types. How
many pounds of the B grade Colombian coffee and how many pounds of the A grade
Arabica coffees are required?
Solution
Let c represent the number of pounds of the B grade Colombian coffee. Then
100 - c equals the number of pounds of the A grade Arabica coffee. See Figure 16.
$5 per pound
$10 per pound
+
Figure 16
B Grade
Colombian
c pounds
+
$7 per pound
=
A Grade
Arabica
100 − c pounds
Blend
=
100 pounds
Since there is to be no difference in revenue between selling the A and B grades
separately and selling the blend, we have
Revenue from B grade
+
Revenue from A grade
=
Revenue from blend
Price per pound Pounds of
Price per pound Pounds of
Price per pound Pounds of
e
fe
f + e
fe
f = e
fe
f
B grade
A grade
blend
of B grade
of A grade
of blend
#
# 1100 - c2 =
#
c
+
+10
+7
100
+5
138
CHAPTER 1 Equations and Inequalities
Now solve the equation:
5c + 101100 - c2
5c + 1000 - 10c
- 5c
c
=
=
=
=
700
700
- 300
60
The manager should blend 60 pounds of B grade Colombian coffee with
100 - 60 = 40 pounds of A grade Arabica coffee to get the desired blend.
Check: The 60 pounds of B grade coffee would sell for 1+52 1602 = +300, and
the 40 pounds of A grade coffee would sell for 1 +102 1402 = +400; the
total revenue, $700, equals the revenue obtained from selling the blend,
as desired.
r
Now Work
PROBLEM
23
4 Solve Uniform Motion Problems
Objects that move at a constant speed are said to be in uniform motion. When the
average speed of an object is known, it can be interpreted as that object’s constant
speed. For example, a bicyclist traveling at an average speed of 25 miles per hour
can be considered in uniform motion with a constant speed of 25 miles per hour.
Uniform Motion Formula
If an object moves at an average speed (rate) r, the distance d covered in
time t is given by the formula
d = rt
(2)
That is, Distance = Rate # Time.
EX AMPLE 5
Physics: Uniform Motion
Tanya, who is a long-distance runner, runs at an average speed of 8 miles per hour
(mi>h). Two hours after Tanya leaves your house, you leave in your Honda and
follow the same route. If your average speed is 40 mi>h, how long will it be before
you catch up to Tanya? How far will each of you be from your home?
Solution
Refer to Figure 17. We use t to represent the time (in hours) that it takes the
Honda to catch up to Tanya. When this occurs, the total time elapsed for Tanya is
t + 2 hours because she left 2 hours earlier.
t50
2h
Time t
t50
Figure 17
Time t
Set up the following table:
Rate
mi/h
Time
h
Distance
mi
Tanya
8
t + 2
8(t + 2)
Honda
40
t
40t
SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
139
The distance traveled is the same for both, which leads to the equation
81t + 22 = 40t
8t + 16 = 40t
32t = 16
t =
1
hour
2
1
hour to catch up to Tanya. Each will have gone 20 miles.
2
1
Check: In 2.5 hours, Tanya travels a distance of 12.52 182 = 20 miles. In hour,
2
1
the Honda travels a distance of a b 1402 = 20 miles.
2
It will take the Honda
r
EXAMPL E 6
Physics: Uniform Motion
A motorboat heads upstream a distance of 24 miles on a river whose current is running
at 3 miles per hour (mi/h). The trip up and back takes 6 hours. Assuming that the
motorboat maintained a constant speed relative to the water, what was its speed?
Solution
See Figure 18. Use r to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is r - 3 mi/h, and the true
speed going downstream is r + 3 mi/h. Since Distance = Rate # Time, then
Distance
Time =
. Set up a table.
Rate
24 miles
Rate
mi/h
Distance
mi
Distance
Time
=
Rate
h
Upstream
r - 3
24
24
r - 3
Downstream
r + 3
24
24
r + 3
r 2 3 mi/h
r 1 3 mi/h
Figure 18
The total time up and back is 6 hours, which gives the equation
24
24
+
= 6
r - 3
r + 3
241r + 32 + 241r - 32
= 6
1r - 32 1r + 32
48r
= 6
r - 9
2
48r = 61r 2 - 92
6r 2 - 48r - 54 = 0
r - 8r - 9 = 0
2
1r - 92 1r + 12 = 0
r = 9 or r = - 1
Add the quotients on the left.
Simplify.
Multiply both sides by r2 - 9.
Place in standard form.
Divide by 6.
Factor.
Apply the Zero-Product Property and solve.
Discard the solution r = - 1 mi/h and conclude that the speed of the motorboat
relative to the water is 9 mi/h.
r
Now Work
PROBLEM
29
140
CHAPTER 1 Equations and Inequalities
5 Solve Constant Rate Job Problems
Here we look at jobs that are performed at a constant rate. The assumption is that
1
if a job can be done in t units of time, then of the job is done in 1 unit of time. In
1 t
other words, if a job takes 4 hours, then of the job is done in 1 hour.
4
EX AMPLE 7
Solution
Working Together to Do a Job
At 10 am Danny is asked by his father to weed the garden. From past experience,
Danny knows that this will take him 4 hours, working alone. His older brother Mike,
when it is his turn to do this job, requires 6 hours. Since Mike wants to go golfing
with Danny and has a reservation for 1 pm, he agrees to help Danny. Assuming no
gain or loss of efficiency, when will they finish if they work together? Can they make
the golf date?
1
1
Set up Table 2. In 1 hour, Danny does of the job, and in 1 hour, Mike does of the
4
6
job. Let t be the time (in hours) that it takes them to do the job together. In 1 hour,
1
then, of the job is completed. Reason as follows:
t
a
Table 2
Hours to
Do Job
Part of
Job Done
in 1 Hour
Danny
4
1
4
Mike
6
1
6
Together
t
1
t
Part done by Danny
Part done by Mike
Part done together
b + a
b = a
b
in 1 hour
in 1 hour
in 1 hour
From Table 2,
1
1
+
4
6
3
2
+
12
12
5
12
5t
1
t
1
=
t
1
=
t
= 12
=
t =
12
5
The model.
LCD = 12 on the left.
Simplify.
Multiply both sides by 12t.
Divide each side by 5.
12
hours, or 2 hours, 24 minutes.
5
They should make the golf date, since they will finish at 12:24 pm.
Working together, Mike and Danny can do the job in
r
Now Work
PROBLEM
35
1.7 Assess Your Understanding
Concepts and Vocabulary
1. The process of using variables to represent unknown
quantities and then finding relationships that involve these
variables is referred to as
.
2. The money paid for the use of money is
.
3. Objects that move at a constant speed are said to be in
.
4. True or False The amount charged for the use of principal
for a given period of time is called the rate of interest.
5. True or False If an object moves at an average speed r,
the distance d covered in time t is given by the formula
d = rt.
6. Suppose that you want to mix two coffees in order to obtain
100 pounds of a blend. If x represents the number of pounds
of coffee A, write an algebraic expression that represents the
number of pounds of coffee B.
(a) 100 - x
(b) x - 100
(c) 100 x
(d) 100 + x
7. Which of the following is the simple interest formula?
P
rt
(b) I = Prt (c) I =
(d) I = P + rt
(a) I =
P
rt
8. If it takes 5 hours to complete a job, what fraction of the job
is done in 1 hour?
5
1
1
4
(a)
(b)
(c)
(d)
5
4
5
4
SECTION 1.7 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
141
Applications and Extensions
In Problems 9–18, translate each sentence into a mathematical equation. Be sure to identify the meaning of all symbols.
9. Geometry The area of a circle is the product of the number
p and the square of the radius.
10. Geometry The circumference of a circle is the product of
the number p and twice the radius.
11. Geometry The area of a square is the square of the length of
a side.
12. Geometry The perimeter of a square is four times the length
of a side.
13. Physics Force equals the product of mass and acceleration.
14. Physics Pressure is force per unit area.
15. Physics Work equals force times distance.
16. Physics Kinetic energy is one-half the product of the mass
and the square of the velocity.
17. Business
The total variable cost of manufacturing
x dishwashers is $150 per dishwasher times the number of
dishwashers manufactured.
18. Business
The total revenue derived from selling
x dishwashers is $250 per dishwasher times the number of
dishwashers sold.
25. Business: Mixing Nuts A nut store normally sells cashews
for $9.00 per pound and almonds for $3.50 per pound. But
at the end of the month the almonds had not sold well, so,
in order to sell 60 pounds of almonds, the manager decided
to mix the 60 pounds of almonds with some cashews and
sell the mixture for $7.50 per pound. How many pounds of
cashews should be mixed with the almonds to ensure no
change in the revenue?
26. Business: Mixing Candy A candy store sells boxes of candy
containing caramels and cremes. Each box sells for $12.50
and holds 30 pieces of candy (all pieces are the same size). If
the caramels cost $0.25 to produce and the cremes cost $0.45
to produce, how many of each should be in a box to yield a
profit of $3?
27. Physics: Uniform Motion A motorboat can maintain a
constant speed of 16 miles per hour relative to the water.
The boat travels upstream to a certain point in 20 minutes;
the return trip takes 15 minutes. What is the speed of the
current? See the figure.
19. Financial Planning Betsy, a recent retiree, requires $6000
per year in extra income. She has $50,000 to invest and
can invest in B-rated bonds paying 15% per year or in a
certificate of deposit (CD) paying 7% per year. How much
money should Betsy invest in each to realize exactly $6000
in interest per year?
20. Financial Planning After 2 years, Betsy (see Problem 19)
finds that she will now require $7000 per year. Assuming
that the remaining information is the same, how should the
money be reinvested?
21. Banking A bank loaned out $12,000, part of it at the rate of
8% per year and the rest at the rate of 18% per year. If the
interest received totaled $1000, how much was loaned at 8%?
22. Banking Wendy, a loan officer at a bank, has $1,000,000 to
lend and is required to obtain an average return of 18% per
year. If she can lend at the rate of 19% or at the rate of 16%,
how much can she lend at the 16% rate and still meet her
requirement?
23. Blending Teas The manager of a store that specializes in
selling tea decides to experiment with a new blend. She will
mix some Earl Grey tea that sells for $5 per pound with some
Orange Pekoe tea that sells for $3 per pound to get 100 pounds
of the new blend. The selling price of the new blend is to be
$4.50 per pound, and there is to be no difference in revenue
between selling the new blend and selling the other types. How
many pounds of the Earl Grey tea and of the Orange Pekoe tea
are required?
24. Business: Blending Coffee A coffee manufacturer wants to
market a new blend of coffee that sells for $3.90 per pound
by mixing two coffees that sell for $2.75 and $5 per pound,
respectively. What amounts of each coffee should be blended
to obtain the desired mixture?
[Hint: Assume that the total weight of the desired blend is
100 pounds.]
28. Physics: Uniform Motion A motorboat heads upstream
on a river that has a current of 3 miles per hour. The trip
upstream takes 5 hours, and the return trip takes 2.5 hours.
What is the speed of the motorboat? (Assume that the boat
maintains a constant speed relative to the water.)
29. Physics: Uniform Motion A motorboat maintained a
constant speed of 15 miles per hour relative to the water in
going 10 miles upstream and then returning. The total time
for the trip was 1.5 hours. Use this information to find the
speed of the current.
30. Physics: Uniform Motion Two cars enter the Florida
Turnpike at Commercial Boulevard at 8:00 am, each
heading for Wildwood. One car’s average speed is 10 miles
per hour more than the other’s. The faster car arrives at
1
Wildwood at 11:00 am, hour before the other car. What
2
was the average speed of each car? How far did each travel?
31. Moving Walkways The speed of a moving walkway is
typically about 2.5 feet per second. Walking on such a
moving walkway, it takes Karen a total of 40 seconds to
travel 50 feet with the movement of the walkway and then
back again against the movement of the walkway. What is
Karen’s normal walking speed?
Source: Answers.com
36. Working Together on a Job Patrice, by himself, can paint
four rooms in 10 hours. If he hires April to help, they can do
the same job together in 6 hours. If he lets April work alone,
how long will it take her to paint four rooms?
37. Enclosing a Garden A gardener has 46 feet of fencing to be
used to enclose a rectangular garden that has a border 2 feet
wide surrounding it. See the figure.
(a) If the length of the garden is to be twice its width, what
will be the dimensions of the garden?
(b) What is the area of the garden?
(c) If the length and width of the garden are to be the same,
what will be the dimensions of the garden?
(d) What will be the area of the square garden?
2 ft
2 ft
38. Construction A pond is enclosed by a wooden deck that is
3 feet wide. The fence surrounding the deck is 100 feet long.
(a) If the pond is square, what are its dimensions?
(b) If the pond is rectangular and the length of the pond is
to be three times its width, what are its dimensions?
(c) If the pond is circular, what is its diameter?
(d) Which pond has the larger area?
39. Football A tight end can run the 100-yard dash in 12 seconds.
A defensive back can do it in 10 seconds. The tight end
catches a pass at his own 20-yard line with the defensive
back at the 15-yard line. (See the figure-top, right.) If no
other players are nearby, at what yard line will the defensive
back catch up to the tight end?
[Hint: At time t = 0, the defensive back is 5 yards behind
the tight end.]
30
20
TE
DB
10
10
35. Working Together on a Job Trent can deliver his
newspapers in 30 minutes. It takes Lois 20 minutes to do
the same route. How long would it take them to deliver the
newspapers if they worked together?
20
34. Laser Printers It takes an HP LaserJet M451dw laser printer
16 minutes longer to complete an 840-page print job
by itself than it takes an HP LaserJet CP4025dn to
complete the same job by itself. Together the two printers
can complete the job in 15 minutes. How long does it take
each printer to complete the print job alone? What is the
speed of each printer?
Source: Hewlett-Packard
30
33. Tennis A regulation doubles tennis court has an area of
2808 square feet. If it is 6 feet longer than twice its width,
determine the dimensions of the court.
Source: United States Tennis Association
40
32. High-Speed Walkways Toronto’s Pearson International
Airport has a high-speed version of a moving walkway. If
Liam walks while riding this moving walkway, he can travel
280 meters in 60 seconds less time than if he stands still on the
moving walkway. If Liam walks at a normal rate of 1.5 meters
per second, what is the speed of the walkway?
Source: Answers.com
40
CHAPTER 1 Equations and Inequalities
SOUTH
142
40. Computing Business Expense
Therese, an outside
salesperson, uses her car for both business and pleasure. Last
year, she traveled 30,000 miles, using 900 gallons of gasoline.
Her car gets 40 miles per gallon on the highway and 25 in
the city. She can deduct all highway travel, but no city travel,
on her taxes. How many miles should Therese deduct as a
business expense?
41. Mixing Water and Antifreeze How much water should be
added to 1 gallon of pure antifreeze to obtain a solution that
is 60% antifreeze?
42. Mixing Water and Antifreeze The cooling system of a
certain foreign-made car has a capacity of 15 liters. If the
system is filled with a mixture that is 40% antifreeze, how
much of this mixture should be drained and replaced by
pure antifreeze so that the system is filled with a solution
that is 60% antifreeze?
43. Chemistry: Salt Solutions How much water must be
evaporated from 32 ounces of a 4% salt solution to make a
6% salt solution?
44. Chemistry: Salt Solutions How much water must be
evaporated from 240 gallons of a 3% salt solution to
produce a 5% salt solution?
45. Purity of Gold The purity of gold is measured in karats,
with pure gold being 24 karats. Other purities of gold are
expressed as proportional parts of pure gold. Thus, 18-karat
18
12
gold is
, or 75% pure gold; 12-karat gold is
, or 50%
24
24
pure gold; and so on. How much 12-karat gold should be
mixed with pure gold to obtain 60 grams of 16-karat gold?
46. Chemistry: Sugar Molecules A sugar molecule has twice as
many atoms of hydrogen as it does oxygen and one more
atom of carbon than of oxygen. If a sugar molecule has a
total of 45 atoms, how many are oxygen? How many are
hydrogen?
47. Running a Race Mike can run the mile in 6 minutes, and
Dan can run the mile in 9 minutes. If Mike gives Dan a head
start of 1 minute, how far from the start will Mike pass Dan?
How long does it take? See the figure.
Dan
Mike
Start
1–
4
1–
2
mi
mi
3–
4
mi
Chapter Review
48. Range of an Airplane An air rescue plane averages
300 miles per hour in still air. It carries enough fuel for 5
hours of flying time. If, upon takeoff, it encounters a head
wind of 30 mi/h, how far can it fly and return safely?
(Assume that the wind remains constant.)
49. Emptying Oil Tankers An oil tanker can be emptied by the
main pump in 4 hours. An auxiliary pump can empty the
tanker in 9 hours. If the main pump is started at 9 am, when
should the auxiliary pump be started so that the tanker is
emptied by noon?
50. Cement Mix A 20-pound bag of Economy brand cement
mix contains 25% cement and 75% sand. How much pure
cement must be added to produce a cement mix that is 40%
cement?
51. Filling a Tub A bathroom tub will fill in 15 minutes with
both faucets open and the stopper in place. With both
faucets closed and the stopper removed, the tub will empty
in 20 minutes. How long will it take for the tub to fill if both
faucets are open and the stopper is removed?
52. Using Two Pumps A 5-horsepower (hp) pump can empty a
pool in 5 hours. A smaller, 2-hp pump empties the same pool
in 8 hours. The pumps are used together to begin emptying
this pool. After two hours, the 2-hp pump breaks down. How
long will it take the larger pump to finish emptying the pool?
53. A Biathlon Suppose that you have entered an 87-mile
biathlon that consists of a run and a bicycle race. During
143
your run, your average speed is 6 miles per hour, and during
your bicycle race, your average speed is 25 miles per hour.
You finish the race in 5 hours. What is the distance of the
run? What is the distance of the bicycle race?
54. Cyclists Two cyclists leave a city at the same time, one going
east and the other going west. The westbound cyclist bikes
5 mph faster than the eastbound cyclist. After 6 hours they
are 246 miles apart. How fast is each cyclist riding?
55. Comparing Olympic Heroes In the 2012 Olympics, Usain
Bolt of Jamaica won the gold medal in the 100-meter race
with a time of 9.69 seconds. In the 1896 Olympics, Thomas
Burke of the United States won the gold medal in the
100-meter race in 12.0 seconds. If they ran in the same race,
repeating their respective times, by how many meters would
Bolt beat Burke?
56. Constructing a Coffee Can A 39-ounce can of Hills Bros.®
coffee requires 188.5 square inches of aluminum. If its height
is 7 inches, what is its radius? [Hint: The surface area S of a
right cylinder is S = 2pr 2 + 2prh, where r is the radius and
h is the height.]
7 in.
39 oz.
Explaining Concepts: Discussion and Writing
57. Critical Thinking You are the manager of a clothing store
and have just purchased 100 dress shirts for $20.00 each.
After 1 month of selling the shirts at the regular price, you
plan to have a sale giving 40% off the original selling price.
However, you still want to make a profit of $4 on each shirt
at the sale price. What should you price the shirts at initially
to ensure this? If, instead of 40% off at the sale, you give
50% off, by how much is your profit reduced?
60. Computing Average Speed In going from Chicago to
Atlanta, a car averages 45 miles per hour, and in going from
Atlanta to Miami, it averages 55 miles per hour. If Atlanta
is halfway between Chicago and Miami, what is the average
speed from Chicago to Miami? Discuss an intuitive solution.
Write a paragraph defending your intuitive solution. Then
solve the problem algebraically. Is your intuitive solution the
same as the algebraic one? If not, find the flaw.
58. Critical Thinking Make up a word problem that requires
solving a linear equation as part of its solution. Exchange
problems with a friend. Write a critique of your friend’s problem.
61. Speed of a Plane On a recent flight from Phoenix to
Kansas City, a distance of 919 nautical miles, the plane arrived
20 minutes early. On leaving the aircraft, I asked the captain,
“What was our tail wind?” He replied, “I don’t know, but our
ground speed was 550 knots.” Has enough information been
provided for you to find the tail wind? If possible, find the
tail wind. (1 knot = 1 nautical mile per hour)
59. Critical Thinking Without solving, explain what is wrong
with the following mixture problem: How many liters of
25% ethanol should be added to 20 liters of 48% ethanol
to obtain a solution of 58% ethanol? Now go through an
algebraic solution. What happens?
Chapter Review
Things to Know
Quadratic formula (pp. 97 and 110)
- b { 2b2 - 4ac
.
2a
If b2 - 4ac 6 0, there are no real solutions.
If ax2 + bx + c = 0, a ≠ 0, then x =
Discriminant (pp. 97 and 110)
If b2 - 4ac 7 0, there are two unequal real solutions.
If b2 - 4ac = 0, there is one repeated real solution, a double root.
If b2 - 4ac 6 0, there are no real solutions, but there are two distinct complex solutions that are not real; the complex solutions
are conjugates of each other.
144
CHAPTER 1 Equations and Inequalities
Interval notation (p. 120)
3a, b4
5x a … x … b6
3a, q 2
5x x Ú a6
1a, b4
5x a 6 x … b6
1 - q , a4
5x x … a6
3a, b2
5x a … x 6 b6
1a, b2
1a, q 2
1 - q , a2
5x a 6 x 6 b6
1 - q, q 2
5x x 7 a6
5x x 6 a6
All real numbers
Properties of inequalities
Addition property (p. 121)
If a 6 b, then a + c 6 b + c.
Multiplication properties (p. 122)
(a) If a 6 b and if c 7 0, then ac 6 bc.
(b) If a 7 b and if c 7 0, then ac 7 bc.
If a 6 b and if c 6 0, then ac 7 bc.
If a 7 b and if c 6 0, then ac 6 bc.
If a 7 b, then a + c 7 b + c.
Reciprocal properties (p. 123)
If a 7 0, then
If
1
7 0.
a
1
7 0, then a 7 0.
a
If a 6 0, then
If
1
6 0.
a
1
6 0, then a 6 0.
a
Absolute value
If 0 u 0 = a, a 7 0, then u = - a or u = a. (p. 130)
If 0 u 0 … a, a 7 0, then - a … u … a. (p. 131)
If 0 u 0 Ú a, a 7 0, then u … - a or u Ú a. (p. 132)
Objectives
Section
1.1
1
2
3
1.2
1
2
3
4
1.3
1
2
1.4
1
2
3
1.5
1
2
3
4
1.6
1
2
1.7
1
2
3
4
5
You should be able to . . .
Examples
Review Exercises
Solve a linear equation (p. 84)
Solve equations that lead to linear equations (p. 86)
Solve problems that can be modeled by linear equations (p. 87)
1–3
4–6
8, 9
1–3, 7, 8
4
45, 57
Solve a quadratic equation by factoring (p. 93)
Solve a quadratic equation by completing the square (p. 95)
Solve a quadratic equation using the quadratic formula (p. 96)
Solve problems that can be modeled by quadratic equations (p. 99)
1, 2
4, 5
6–9
10
6, 9, 21, 22
5, 6, 9, 10, 13, 21, 22
5, 6, 9, 10, 13, 21, 22
50, 54, 56
Add, subtract, multiply, and divide complex numbers (p. 105)
Solve quadratic equations in the complex number system (p. 109)
1–8
9–12
35–39
40–43
Solve radical equations (p. 113)
Solve equations quadratic in form (p. 114)
Solve equations by factoring (p. 116)
1–3
4–6
7, 8
11, 12, 15–19, 23
14, 20
26, 27
Use interval notation (p. 120)
Use properties of inequalities (p. 121)
Solve inequalities (p. 123)
Solve combined inequalities (p. 124)
1, 2
3–6
7, 8
9, 10
28–34
28–34, 47
28, 47
29, 30
Solve equations involving absolute value (p. 130)
Solve inequalities involving absolute value (p. 130)
1
2–6
24, 25
31–34
Translate verbal descriptions into mathematical expressions (p. 135)
Solve interest problems (p. 136)
Solve mixture problems (p. 137)
Solve uniform motion problems (p. 138)
Solve constant rate job problems (p. 140)
1
2, 3
4
5, 6
7
44
45
53, 55
46, 48, 49, 59, 60
51, 52, 58
Chapter Review
145
Review Exercises
In Problems 1–27, find the real solutions, if any, of each equation. (Where they appear, a, b, m, and n are positive constants.)
1. 2 -
x
= 8
3
4.
x
6
=
x - 1
5
7.
1
1
3
x
ax - b = 2
3
4
6
x ≠ 1
x
1
3x
- =
4
3
12
2. - 215 - 3x2 + 8 = 4 + 5x
3.
5. x11 - x2 = 6
6. x11 + x2 = 6
8.
9. 1x - 12 12x + 32 = 3
1 - 3x
x + 6
1
=
+
4
3
2
10. 2x + 3 = 4x2
3 2
11. 2
x - 1 = 2
12. 21 + x3 = 3
13. x1x + 12 + 2 = 0
14. x4 - 5x2 + 4 = 0
15. 22x - 3 + x = 3
4
16. 2
2x + 3 = 2
17. 2x + 1 + 2x - 1 = 22x + 1
18. 22x - 1 - 2x - 5 = 3
19. 2x1>2 - 3 = 0
20. x -6 - 7x -3 - 8 = 0
21. x2 + m2 = 2mx + 1nx2 2 n ≠ 1
23. 2x2 + 3x + 7 - 2x2 - 3x + 9 + 2 = 0
22. 10a2 x2 - 2abx - 36b2 = 0
24. 0 2x + 3 0 = 7
25. 0 2 - 3x 0 + 2 = 9
26. 2x3 = 3x2
27. 2x3 + 5x2 - 8x - 20 = 0
In Problems 28–34, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.
28.
2x - 3
x
+ 2 …
5
2
32. 0 2x - 5 0 Ú 9
29. - 9 …
2x + 3
… 7
-4
33. 2 + 0 2 - 3x 0 … 4
30. 2 6
31. 0 3x + 4 0 6
3 - 3x
6 6
12
1
2
34. 1 - 0 2 - 3x 0 6 - 4
In Problems 35–39, use the complex number system and write each expression in the standard form a + bi.
35. 16 + 3i2 - 12 - 4i2
36. 413 - i2 + 31 - 5 + 2i2
38. i 50
39. 12 + 3i2 3
37.
3
3 + i
In Problems 40–43, solve each equation in the complex number system.
40. x2 + x + 1 = 0
41. 2x2 + x - 2 = 0
44. Translate the following statement into a mathematical
expression: The total cost C of manufacturing x bicycles in
one day is $50,000 plus $95 times the number of bicycles
manufactured.
45. Financial Planning Steve, a recent retiree, requires $5000
per year in extra income. He has $70,000 to invest and
can invest in A-rated bonds paying 8% per year or in a
certificate of deposit (CD) paying 5% per year. How much
money should be invested in each to realize exactly $5000 in
interest per year?
46. Lightning and Thunder A flash of lightning is seen, and the
resulting thunderclap is heard 3 seconds later. If the speed
of sound averages 1100 feet per second, how far away is the
storm?
42. x2 + 3 = x
43. x11 - x2 = 6
146
CHAPTER 1 Equations and Inequalities
47. Physics: Intensity of Light The intensity I (in candlepower)
900
of a certain light source obeys the equation I = 2 , where
x
x is the distance (in meters) from the light. Over what range
of distances can an object be placed from this light source
so that the range of intensity of light is from 1600 to 3600
candlepower, inclusive?
53. Chemistry: Salt Solutions How much water should be
added to 64 ounces of a 10% salt solution to make a 2% salt
solution?
54. Geometry The diagonal of a rectangle measures 10 inches.
If the length is 2 inches more than the width, find the
dimensions of the rectangle.
55. Chemistry: Mixing Acids A laboratory has 60 cubic
centimeters (cm3) of a solution that is 40% HCl acid. How
many cubic centimeters of a 15% solution of HCl acid
should be mixed with the 60 cm3 of 40% acid to obtain a
solution of 25% HCl? How much of the 25% solution is there?
48. Extent of Search and Rescue A search plane has a cruising
speed of 250 miles per hour and carries enough fuel for at
most 5 hours of flying. If there is a wind that averages 30
miles per hour and the direction of the search is with the
wind one way and against it the other, how far can the search
plane travel before it has to turn back?
56. Framing a Painting An artist has 50 inches of oak trim to
frame a painting. The frame is to have a border 3 inches wide
surrounding the painting.
(a) If the painting is square, what are its dimensions? What
are the dimensions of the frame?
(b) If the painting is rectangular with a length twice its
width, what are the dimensions of the painting? What
are the dimensions of the frame?
49. Rescue at Sea A life raft, set adrift from a sinking ship
150 miles offshore, travels directly toward a Coast Guard
station at the rate of 5 miles per hour. At the time that the
raft is set adrift, a rescue helicopter is dispatched from the
Coast Guard station. If the helicopter’s average speed is
90 miles per hour, how long will it take the helicopter to
reach the life raft?
57. Finance An inheritance of $900,000 is to be divided among
Scott, Alice, and Tricia in the following manner: Alice is to
1
3
receive of what Scott gets, while Tricia gets of what Scott
4
2
gets. How much does each receive?
90 mi/h
58. Utilizing Copying Machines A new copying machine can
do a certain job in 1 hour less than an older copier. Together
they can do this job in 72 minutes. How long would it take
the older copier by itself to do the job?
5 mi/h
150 mi
59. Evening Up a Race In a 100-meter race, Todd crosses the
finish line 5 meters ahead of Scott. To even things up, Todd
suggests to Scott that they race again, this time with Todd
lining up 5 meters behind the start.
(a) Assuming that Todd and Scott run at the same pace as
before, does the second race end in a tie?
(b) If not, who wins?
50. Physics An object is thrown down from the top of a building
1280 feet tall with an initial velocity of 32 feet per second.
The distance s (in feet) of the object from the ground after
t seconds is s = 1280 - 32t - 16t 2.
(a) When will the object strike the ground?
(b) What is the height of the object after 4 seconds?
(c) By how many meters does he win?
(d) How far back should Todd start so that the race ends in
a tie?
51. Working Together to Get a Job Done Clarissa and Shawna,
working together, can paint the exterior of a house in 6 days.
Clarissa by herself can complete this job in 5 days less than
Shawna. How long will it take Clarissa to complete the job
by herself?
After running the race a second time, Scott, to even things
up, suggests to Todd that he (Scott) line up 5 meters in front
of the start.
(e) Assuming again that they run at the same pace as in the
first race, does the third race result in a tie?
(f) If not, who wins?
(g) By how many meters?
(h) How far ahead should Scott start so that the race ends in
a tie?
52. Emptying a Tank Two pumps of different sizes, working
together, can empty a fuel tank in 5 hours. The larger pump
can empty this tank in 4 hours less than the smaller one. If
the larger pump is out of order, how long will it take the
smaller one to do the job alone?
60. Physics: Uniform Motion A man is walking at an average speed of 4 miles per hour alongside a railroad track. A freight train, going
in the same direction at an average speed of 30 miles per hour, requires 5 seconds to pass the man. How long is the freight train?
Give your answer in feet.
30 mi/h
Length of train
4 mi/h
t50
5 sec
t55
Chapter Projects
The Chapter Test Prep Videos are step-by-step solutions available in
, or on this text’s
Channel. Flip back to the Resources
for Success page for a link to this text’s YouTube channel.
Chapter Test
In Problems 1–7, solve each equation.
2x
x
5
2. x1x - 12 = 6
1.
- =
3
2
12
5. 0 2x - 3 0 + 7 = 10
147
6. 3x3 + 2x2 - 12x - 8 = 0
4. 22x - 5 + 2 = 4
3. x4 - 3x2 - 4 = 0
7. 3x2 - x + 1 = 0
In Problems 8–10, solve each inequality. Express your answer using interval notation. Graph the solution set.
8. - 3 …
11. Write
3x - 4
… 6
2
9. 0 3x + 4 0 6 8
10. 2 + 0 2x - 5 0 Ú 9
-2
in the standard form a + bi.
3 - i
12. Solve the equation 4x2 - 4x + 5 = 0 in the complex number system.
13. Blending Coffee A coffee house has 20 pounds of a coffee that sells for $4 per pound. How many pounds of a coffee that sells for
$8 per pound should be mixed with the 20 pounds of $4-per-pound coffee to obtain a blend that will sell for $5 per pound? How
much of the $5-per-pound coffee is there to sell?
Chapter Projects
time t, measured in months, is the length of the loan. For
example, a 30-year loan requires 12 * 30 = 360 monthly
payments.
P = L≥
Internet-based Project
I. Financing a Purchase At some point in your life, you are
likely to need to borrow money to finance a purchase.
For example, most of us will finance the purchase of a
car or a home. What is the mathematics behind financing
a purchase? When you borrow money from a bank,
the bank uses a rather complex equation (or formula) to
determine how much you need to pay each month to repay
the loan. There are a number of variables that determine
the monthly payment. These variables include the amount
borrowed, the interest rate, and the length of the loan. The
interest rate is based on current economic conditions, the
length of the loan, the type of item being purchased, and
your credit history.
The following formula gives the monthly payment P
required to pay off a loan amount L at an annual interest rate r,
expressed as a decimal, but usually given as a percent. The
r
12
1 - a1 +
r -t
b
12
¥
P = monthly payment
L = loan amount
r = annual rate of interest
expressed as a decimal
t = length of loan, in months
1. Interest rates change daily. Many websites post current
interest rates on loans. Go to www.bankrate.com (or
some other website that posts lenders’ interest rates) and
find the current best interest rate on a 60-month new-car
purchase loan. Use this rate to determine the monthly
payment on a $30,000 automobile loan.
2. Determine the total amount paid for the loan by
multiplying the loan payment by the term of the loan.
3. Determine the total amount of interest paid by subtracting
the loan amount from the total amount paid from
question 2.
4. More often than not, we decide how much of a payment
we can afford and use that information to determine
the loan amount. Suppose you can afford a monthly
payment of $500. Use the interest rate from question 1 to
determine the maximum amount you can borrow. If you
have $5000 to put down on the car, what is the maximum
value of a car you can purchase?
5. Repeat questions 1 through 4 using a 72-month new-car
purchase loan, a 60-month used-car purchase loan, and a
72-month used-car purchase loan.
148
CHAPTER 1 Equations and Inequalities
6. We can use the power of a spreadsheet, such as Excel, to create a loan amortization schedule. A loan amortization schedule is a
list of the monthly payments, a breakdown of interest and principal, along with a current loan balance. Create a loan amortization
schedule for each of the four loan scenarios discussed on the previous page, using the following as a guide. You may want to use
an Internet search engine to research specific keystrokes for creating an amortization schedule in a spreadsheet. We supply a
sample spreadsheet with formulas included as a guide. Use the spreadsheet to verify your results from questions 1 through 5.
Loan Information
Loan Amount
Annual Interest Rate
Length of Loan (years)
Number of Payments
7.
$30,000.00
0.045
5
=B4*12
Payment
Number
Payment Amount
Interest
Principal
Balance
Total Interest
Paid
1
2
3
·
·
·
=PMT($B$3/12,$B$5,−$B$2,0)
=PMT($B$3/12,$B$5,−$B$2,0)
=PMT($B$3/12,$B$5,−$B$2,0)
·
·
·
=B2*$B$3/12
=H2*$B$3/12
=H3*$B$3/12
·
·
·
=E2−F2
=E3−F3
=E4−F4
·
·
·
=B2−G2
=H2−G3
=H3−G4
·
·
·
=B2*$B$3/12
=I2+F3
=I3+F4
·
·
·
Go to an online automobile website such as www.cars.com, www.edmunds.com, or www.autobytel.com. Research the types
of vehicles you can afford for a monthly payment of $500. Decide on a vehicle you would purchase based on your analysis in
questions 1–6. Be sure to justify your decision, and include the impact the term of the loan has on your decision. You might
consider other factors in your decision, such as expected maintenance costs and insurance costs.
Citation: Excel ©2013 Microsoft Corporation. Used with permission from Microsoft.
The following project is also available on the Instructor’s Resource Center (IRC):
II. Project at Motorola How Many Cellular Phones Can I Make? An industrial engineer uses a model involving equations to be sure
production levels meet customer demand.
2
Graphs
How to Value a House
Two things to consider in valuing a home are, first, how does it compare to similar
homes that have sold recently? Is the asking price fair? And second, what value do
you place on the advertised features and amenities? Yes, other people might
value them highly, but do you?
Zestimate home valuation, RealEstateABC.com, and Reply.com are
among the many algorithmic (generated by a computer model) starting
points in figuring out the value of a home. They show you how the home is
priced relative to other homes in the area, but you need to add in all the things
that only someone who has seen the house knows. You can do that using
My Estimator, and then you create your own estimate and see how it stacks
up against the asking price.
Looking at “Comps”
Knowing whether an asking price is fair will be important when you’re ready to make
an offer on a house. It will be even more important when your mortgage lender hires
an appraiser to determine whether the house is worth the loan you want.
Check with your agent, Zillow.com, propertyshark.com, or other websites to see
recent sales of homes in the area that are similar, or comparable, to what you’re looking
for. Print them out and keep these “comps” in a three-ring binder; you’ll be referring
to them quite a bit.
Note that “recent sales” usually means within the last six months. A sales price
from a year ago may bear little or no relation to what is going on in your area right
now. In fact, some lenders will not accept comps older than three months.
Market activity also determines how easy or difficult it is to find accurate comps.
In a “hot” or busy market, with sales happening all the time, you’re likely to have
lots of comps to choose from. In a less active market, finding reasonable comps
becomes harder. And if the home you’re looking at has special design features, finding a
comparable property is harder still. It’s also necessary to know what’s going on in a
given sub-segment. Maybe large, high-end homes are selling like hotcakes, but owners
of smaller houses are staying put, or vice versa.
Source: http://allmyhome.blogspot.com/2008/07/how-to-value-house.html
—See the Internet-based Chapter Project—
A Look Back
Chapter R and Chapter 1 review skills from intermediate algebra.
A Look Ahead
Here we connect algebra and geometry using the rectangular coordinate
system. In the 1600s, algebra had developed to the point that René Descartes
(1596–1650) and Pierre de Fermat (1601–1665) were able to use rectangular
coordinates to translate geometry problems into algebra problems, and vice versa.
This enabled both geometers and algebraists to gain new insights into their subjects,
which had been thought to be separate but now were seen as connected.
Outline
2.1
2.2
2.3
2.4
2.5
The Distance and Midpoint Formulas
Graphs of Equations in Two Variables;
Intercepts; Symmetry
Lines
Circles
Variation
Chapter Review
Chapter Test
Cumulative Review
Chapter Project
149
149
150
CHAPTER 2 Graphs
2.1 The Distance and Midpoint Formulas
PREPARING FOR THIS SECTION Before getting started, review the following:
r Algebra Essentials (Chapter R, Section R.2,
pp. 17–26)
r Geometry Essentials (Chapter R, Section R.3,
pp. 30–35)
Now Work the ‘Are You Prepared?’ problems on page 154.
OBJECTIVES 1 Use the Distance Formula (p. 151)
2 Use the Midpoint Formula (p. 153)
Rectangular Coordinates
y
4
2
–4
–2
O
2
4
x
–2
–4
Figure 1 xy@Plane
y
4
3
(–3, 1)
1
–4
3
(–2, –3)
3
(3, 2)
2
O
3
x
4
2
(3, –2)
2
Figure 2
y
Quadrant II
x < 0, y > 0
Quadrant I
x > 0, y > 0
Quadrant III
x < 0, y < 0
Quadrant IV
x > 0, y < 0
x
Figure 3
We locate a point on the real number line by assigning it a single real number, called
the coordinate of the point. For work in a two-dimensional plane, we locate points
by using two numbers.
Begin with two real number lines located in the same plane: one horizontal and
the other vertical. The horizontal line is called the x-axis, the vertical line the y-axis,
and the point of intersection the origin O. See Figure 1. Assign coordinates to every
point on these number lines using a convenient scale. In mathematics, we usually
use the same scale on each axis, but in applications, different scales appropriate to
the application may be used.
The origin O has a value of 0 on both the x-axis and the y-axis. Points on the
x-axis to the right of O are associated with positive real numbers, and those to the
left of O are associated with negative real numbers. Points on the y-axis above O
are associated with positive real numbers, and those below O are associated with
negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y,
respectively, and an arrow at the end of each axis is used to denote the positive
direction.
The coordinate system described here is called a rectangular or Cartesian*
coordinate system. The plane formed by the x-axis and y-axis is sometimes called
the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes.
Any point P in the xy-plane can be located by using an ordered pair 1x, y2 of
real numbers. Let x denote the signed distance of P from the y-axis (signed means
that if P is to the right of the y-axis, then x 7 0, and if P is to the left of the y-axis,
then x 6 0); and let y denote the signed distance of P from the x-axis. The ordered
pair 1x, y2 , also called the coordinates of P, gives us enough information to locate
the point P in the plane.
For example, to locate the point whose coordinates are 1 - 3, 12, go 3 units
along the x-axis to the left of O and then go straight up 1 unit. We plot this point
by placing a dot at this location. See Figure 2, in which the points with coordinates
1 - 3, 12 , 1 - 2, - 32, 13, - 22 , and 13, 22 are plotted.
The origin has coordinates 10, 02 . Any point on the x-axis has coordinates of
the form 1x, 02, and any point on the y-axis has coordinates of the form 10, y2 .
If 1x, y2 are the coordinates of a point P, then x is called the x-coordinate, or
abscissa, of P, and y is the y-coordinate, or ordinate, of P. We identify the point P by
its coordinates 1x, y2 by writing P = 1x, y2 . Usually, we will simply say “the point
1x, y2 ” rather than “the point whose coordinates are 1x, y2 .”
The coordinate axes divide the xy-plane into four sections called quadrants, as
shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all
points are positive; in quadrant II, x is negative and y is positive; in quadrant III,
both x and y are negative; and in quadrant IV, x is positive and y is negative. Points
on the coordinate axes belong to no quadrant.
Now Work
PROBLEM
15
* Named after René Descartes (1596–1650), a French mathematician, philosopher, and theologian.
SECTION 2.1 The Distance and Midpoint Formulas
151
COMMENT On a graphing calculator, you can set the scale on each axis. Once this has been done,
you obtain the viewing rectangle. See Figure 4 for a typical viewing rectangle. You should now read
Section 1, The Viewing Rectangle, in the Appendix.
Figure 4 TI-84 Plus C Standard Viewing Rectangle
■
1 Use the Distance Formula
If the same units of measurement (such as inches, centimeters, and so on) are used
for both the x-axis and y-axis, then all distances in the xy-plane can be measured
using this unit of measurement.
EXAM PL E 1
Solution
Finding the Distance between Two Points
Find the distance d between the points 11, 32 and 15, 62 .
First plot the points 11, 32 and 15, 62 and connect them with a straight line. See
Figure 5(a). To find the length d, begin by drawing a horizontal line from 11, 32 to
15, 32 and a vertical line from 15, 32 to 15, 62, forming a right triangle, as shown in
Figure 5(b). One leg of the triangle is of length 4 (since 0 5 - 1 0 = 4), and the other
is of length 3 (since 0 6 - 3 0 = 3). By the Pythagorean Theorem, the square of the
distance d that we seek is
d 2 = 42 + 32 = 16 + 9 = 25
d = 225 = 5
y
6
y
6
(5, 6)
d
d
3
(5, 6)
3
(1, 3)
6 x
3
3
(1, 3) 4 (5, 3)
3
6 x
(b)
(a)
Figure 5
r
The distance formula provides a straightforward method for computing the
distance between two points.
THEOREM
Distance Formula
The distance between two points P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2, denoted by
d 1P1 , P2 2, is
In Words
To compute the distance between
two points, find the difference
of the x-coordinates, square it,
and add this to the square of the
difference of the y-coordinates.
The square root of this sum is the
distance.
d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2
(1)
Proof of the Distance Formula Let 1x1 , y1 2 denote the coordinates of point P1
and let 1x2 , y2 2 denote the coordinates of point P2. Assume that the line joining
P1 and P2 is neither horizontal nor vertical. Refer to Figure 6(a) on page 152. The
coordinates of P3 are 1x2 , y1 2 . The horizontal distance from P1 to P3 is the absolute
152
CHAPTER 2 Graphs
value of the difference of the x-coordinates, 0 x2 - x1 0 . The vertical distance from
P3 to P2 is the absolute value of the difference of the y-coordinates, 0 y2 - y1 0 . See
Figure 6(b). The distance d 1P1 , P2 2 is the length of the hypotenuse of the right
triangle, so, by the Pythagorean Theorem, it follows that
3 d 1P1 , P2 2 4 2 = 0 x2 - x1 0 2 + 0 y2 - y1 0 2
= 1x2 - x1 2 2 + 1y2 - y1 2 2
d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2
y
y
y2
P2 (x2, y2)
y1
P3 (x2, y1)
P1 (x1, y1)
x2
x1
P2 (x2, y2)
y2
d(P1, P2)
y1
P1 (x1, y1)
x
⏐y2 y1⏐
⏐x2 x1⏐
x1
x2
P3 (x2, y1)
x
(b)
(a)
Figure 6
Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1
equals the y-coordinate of P2; that is, y1 = y2 . Refer to Figure 7(a). In this case, the
distance formula (1) still works, because for y1 = y2 , it reduces to
d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 02 = 2 1x2 - x1 2 2 = 0 x2 - x1 0
y
y
y2
P1 (x1, y1)
y1
d (P1, P2)
P2 (x2, y1)
P1 (x1, y1)
y1
⏐x2 x1⏐
x1
P2 (x1, y2)
⏐y2 y1⏐ d (P1, P2)
x2
x1
x
(a)
x
(b)
Figure 7
A similar argument holds if the line joining P1 and P2 is vertical. See
Figure 7(b).
■
EX AMPLE 2
Solution
Using the Distance Formula
Find the distance d between the points 1 - 4, 52 and (3, 2).
Using the distance formula, equation (1), reveals that the distance d is
d = 2 3 3 - 1 - 42 4 2 + 12 - 52 2 = 272 + 1 - 32 2
= 249 + 9 = 258 ≈ 7.62
Now Work
PROBLEMS
19
AND
23
r
The distance between two points P1 = 1x1, y1 2 and P2 = 1x2, y2 2 is never a
negative number. Also, the distance between two points is 0 only when the points are
identical—that is, when x1 = x2 and y1 = y2. And, because 1x2 - x1 2 2 = 1x1 - x2 2 2
and 1y2 - y1 2 2 = 1y1 - y2 2 2, it makes no difference whether the distance is
computed from P1 to P2 or from P2 to P1; that is, d 1P1 , P2 2 = d 1P2 , P1 2 .
The introduction to this chapter mentioned that rectangular coordinates enable
us to translate geometry problems into algebra problems, and vice versa. The next
example shows how algebra (the distance formula) can be used to solve geometry
problems.
SECTION 2.1 The Distance and Midpoint Formulas
153
Using Algebra to Solve Geometry Problems
EXAMPL E 3
Consider the three points A = 1 - 2, 12, B = 12, 32, and C = 13, 12 .
(a)
(b)
(c)
(d)
Solution
Plot each point and form the triangle ABC.
Find the length of each side of the triangle.
Show that the triangle is a right triangle.
Find the area of the triangle.
(a) Figure 8 shows the points A, B, C and the triangle ABC.
(b) To find the length of each side of the triangle, use the distance formula,
equation (1).
d 1A, B2 = 2 3 2 - 1 - 22 4 2 + 13 - 12 2 = 216 + 4 = 220 = 225
y
d 1B, C2 = 2 13 - 22 2 + 11 - 32 2 = 21 + 4 = 25
B = (2, 3)
3
A = (–2, 1)
d 1A, C2 = 2 3 3 - 1 - 22 4 2 + 11 - 12 2 = 225 + 0 = 5
C = (3, 1)
–3
(c) If the sum of the squares of the lengths of two of the sides equals the square of the
length of the third side, then the triangle is a right triangle. Looking at Figure 8,
it seems reasonable to conjecture that the angle at vertex B might be a right
angle. We shall check to see whether
x
3
Figure 8
3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 3 d 1A, C2 4 2
Using the results in part (b) yields
3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 =
1 225 2 2
+
1 25 2 2
= 20 + 5 = 25 = 3 d 1A, C2 4 2
It follows from the converse of the Pythagorean Theorem that triangle ABC is
a right triangle.
(d) Because the right angle is at vertex B, the sides AB and BC form the base and
height of the triangle. Its area is
Area =
1
1
1Base2 1Height2 = 1 225 2 1 25 2 = 5 square units
2
2
Now Work
PROBLEM
r
31
2 Use the Midpoint Formula
y
P2 = (x 2, y2)
y2
M = (x, y)
y
y1
y – y1
x – x1
P1 = (x1, y1)
x1
Figure 9
y2 – y
x2 – x
B = (x 2, y)
x - x1 = x2 - x
A = (x, y1)
x
We now derive a formula for the coordinates of the midpoint of a line segment.
Let P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 be the endpoints of a line segment, and let
M = 1x, y2 be the point on the line segment that is the same distance from P1 as it is
from P2 . See Figure 9. The triangles P1 AM and MBP2 are congruent. [Do you see why?
d 1P1 , M2 = d 1M, P2 2 is given; also, ∠AP1 M = ∠BMP2* and ∠P1 MA = ∠MP2 B.
Thus, we have angle–side–angle.] Because triangles P1 AM and MBP2 are congruent,
corresponding sides are equal in length. That is,
x2
x
2x = x1 + x2
x =
x1 + x2
2
and
y - y1 = y2 - y
2y = y1 + y2
y =
y1 + y2
2
*A postulate from geometry states that the transversal P1P2 forms congruent corresponding angles with
the parallel line segments P1A and MB.
154
CHAPTER 2 Graphs
THEOREM
Midpoint Formula
The midpoint M = 1x, y2 of the line segment from P1 = 1x1 , y1 2 to
P2 = 1x2, y2 2 is
In Words
M = 1x, y2 = ¢
To find the midpoint of a line
segment, average the x-coordinates
of the endpoints, and average the
y-coordinates of the endpoints.
Find the midpoint of the line segment from P1 = 1 - 5, 52 to P2 = 13, 12 . Plot the
points P1 and P2 and their midpoint.
Solution
y
M (–1, 3)
–5
Apply the midpoint formula (2) using x1 = - 5, y1 = 5, x2 = 3, and y2 = 1. Then
the coordinates 1x, y2 of the midpoint M are
x =
5
y1 + y2
x1 + x2
-5 + 3
5 + 1
=
= - 1 and y =
=
= 3
2
2
2
2
That is, M = 1 - 1, 32 . See Figure 10.
P2 (3, 1)
5
(2)
Finding the Midpoint of a Line Segment
EX AMPLE 4
P1 (–5, 5)
x1 + x2 y1 + y2
,
≤
2
2
x
Now Work
PROBLEM
r
37
Figure 10
2.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. On the real number line, the origin is assigned the number
. (p. 17)
2. If - 3 and 5 are the coordinates of two points on the real
number line, the distance between these points is
.
(pp. 19–20)
3. If 3 and 4 are the legs of a right triangle, the hypotenuse is
. (p. 30)
5. The area A of a triangle whose base is b and whose altitude
is h is A =
. (p. 31)
6. True or False Two triangles are congruent if two angles and
the included side of one equals two angles and the included
side of the other. (pp. 32–33).
4. Use the converse of the Pythagorean Theorem to show that
a triangle whose sides are of lengths 11, 60, and 61 is a right
triangle. (pp. 30–31)
Concepts and Vocabulary
7. If 1x, y2 are the coordinates of a point P in the xy-plane,
then x is called the
of P, and y
is the
of P.
8. The coordinate axes divide the xy-plane into four sections
.
called
9. If three distinct points P, Q, and R all lie on a line, and if
d1P, Q2 = d1Q, R2, then Q is called the
of the line segment from P to R.
10. True or False The distance between two points is
sometimes a negative number.
11. True or False The point 1 - 1, 42 lies in quadrant IV of the
Cartesian plane.
12. True or False The midpoint of a line segment is found by
averaging the x-coordinates and averaging the y-coordinates
of the endpoints.
13. Which of the following statements is true for a point (x, y)
that lies in quadrant III?
(a) Both x and y are positive.
(b) Both x and y are negative.
(c) x is positive, and y is negative.
(d) x is negative, and y is positive.
14. Choose the formula that gives the distance between two
points (x1, y1) and (x2, y2).
(a) 2(x2 - x1)2 + (y2 - y1)2
(b) 2(x2 + x1)2 - (y2 + y1)2
(c) 2(x2 - x1)2 - (y2 - y1)2
(d) 2(x2 + x1)2 + (y2 + y1)2
155
SECTION 2.1 The Distance and Midpoint Formulas
Skill Building
In Problems 15 and 16, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies.
15. (a) A = 1 - 3, 22
(b) B = 16, 02
(c) C = 1 - 2, - 22
(d) D = 16, 52
(e) E = 10, - 32
(f) F = 16, - 32
(d) D = 14, 12
(e) E = 10, 12
(f) F = 1 - 3, 02
16. (a) A = 11, 42
(b) B = 1 - 3, - 42
(c) C = 1 - 3, 42
17. Plot the points 12, 02, 12, - 32, 12, 42, 12, 12, and 12, - 12. Describe the set of all points of the form 12, y2, where y is a real number.
18. Plot the points 10, 32, 11, 32, 1 - 2, 32, 15, 32, and 1 - 4, 32. Describe the set of all points of the form 1x, 32, where x is a real number.
In Problems 19–30, find the distance d1P1 , P2 2 between the points P1 and P2 .
y
19.
–2
–1
y
20.
2 P = (2, 1)
2
P1 = (0, 0)
2
x
21.
P2 = (–2, 1) 2 P = (0, 0)
1
–2
–1
2
x
P2 (–2, 2)
–2
y
2
–1
y
22.
P1 = (–1, 1) 2
P1 (1, 1)
2 x
–2
23. P1 = 13, - 42; P2 = 15, 42
24. P1 = 1 - 1, 02; P2 = 12, 42
25. P1 = 1 - 3, 22; P2 = 16, 02
26. P1 = 12, - 32; P2 = 14, 22
27. P1 = 14, - 32; P2 = 16, 42
28. P1 = 1 - 4, - 32; P2 = 16, 22
29. P1 = 1a, b2; P2 = 10, 02
30. P1 = 1a, a2; P2 = 10, 02
P2 = (2, 2)
–1
2
x
In Problems 31–36, plot each point and form the triangle ABC. Show that the triangle is a right triangle. Find its area.
31. A = 1 - 2, 52; B = 11, 32; C = 1 - 1, 02
32. A = 1 - 2, 52; B = 112, 32; C = 110, - 112
33. A = 1 - 5, 32; B = 16, 02; C = 15, 52
34. A = 1 - 6, 32; B = 13, - 52; C = 1 - 1, 52
35. A = 14, - 32; B = 10, - 32; C = 14, 22
36. A = 14, - 32; B = 14, 12; C = 12, 12
In Problems 37–44, find the midpoint of the line segment joining the points P1 and P2 .
37. P1 = 13, - 42; P2 = 15, 42
38. P1 = 1 - 2, 02; P2 = 12, 42
39. P1 = 1 - 3, 22; P2 = 16, 02
40. P1 = 12, - 32; P2 = 14, 22
41. P1 = 14, - 32; P2 = 16, 12
42. P1 = 1 - 4, - 32; P2 = 12, 22
43. P1 = 1a, b2; P2 = 10, 02
44. P1 = 1a, a2; P2 = 10, 02
Applications and Extensions
45. If the point 12, 52 is shifted 3 units to the right and 2 units
down, what are its new coordinates?
46. If the point 1 - 1, 62 is shifted 2 units to the left and 4 units
up, what are its new coordinates?
47. Find all points having an x-coordinate of 3 whose distance
from the point 1 - 2, - 12 is 13.
(a) By using the Pythagorean Theorem.
(b) By using the distance formula.
48. Find all points having a y-coordinate of - 6 whose distance
from the point 11, 22 is 17.
(a) By using the Pythagorean Theorem.
(b) By using the distance formula.
49. Find all points on the x-axis that are 6 units from the point
14, - 32 .
50. Find all points on the y-axis that are 6 units from the point
14, - 32 .
(a) Find the coordinates of the point if A is shifted 3 units to
the left and 4 units down.
(b) Find the coordinates of the point if A is shifted 2 units to
the left and 8 units up.
52. Plot the points A = 1 - 1, 82 and M = 12, 32 in the
xy-plane. If M is the midpoint of a line segment AB, find the
coordinates of B.
53. The midpoint of the line segment from P1 to P2 is 1 - 1, 42 .
If P1 = 1 - 3, 62 , what is P2?
54. The midpoint of the line segment from P1 to P2 is 15, - 42 . If
P2 = 17, - 22, what is P1?
55. Geometry The medians of a triangle are the line segments
from each vertex to the midpoint of the opposite side (see
the figure). Find the lengths of the medians of the triangle
with vertices at A = 10, 02, B = 16, 02 , and C = 14, 42 .
C
51. Suppose that A = 12, 52 are the coordinates of a point in
the xy-plane.
Median
Midpoint
A
B
156
CHAPTER 2 Graphs
56. Geometry An equilateral triangle is one in which all three
sides are of equal length. If two vertices of an equilateral
triangle are 10, 42 and 10, 02, find
the third vertex. How many of these
s
s
triangles are possible?
s
57. Geometry Find the midpoint of each diagonal of a square
with side of length s. Draw the conclusion that the diagonals
of a square intersect at their midpoints.
[Hint: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the
square.]
a 23 a
58. Geometry Verify that the points (0, 0), (a, 0), and a ,
b
2
2
are the vertices of an equilateral triangle. Then show that
the midpoints of the three sides are the vertices of a second
equilateral triangle (refer to Problem 56).
In Problems 59–62, find the length of each side of the triangle
determined by the three points P1 , P2 , and P3. State whether
the triangle is an isosceles triangle, a right triangle, neither of these,
or both. (An isosceles triangle is one in which at least two of the
sides are of equal length.)
59. P1 = 12, 12; P2 = 1 - 4, 12; P3 = 1 - 4, - 32
60. P1 = 1 - 1, 42; P2 = 16, 22; P3 = 14, - 52
61. P1 = 1 - 2, - 12; P2 = 10, 72; P3 = 13, 22
66. Little League Baseball Refer to Problem 64. Overlay a
rectangular coordinate system on a Little League baseball
diamond so that the origin is at home plate, the positive
x-axis lies in the direction from home plate to first base, and
the positive y-axis lies in the direction from home plate to
third base.
(a) What are the coordinates of first base, second base, and
third base? Use feet as the unit of measurement.
(b) If the right fielder is located at 1180, 202, how far is it from
the right fielder to second base?
(c) If the center fielder is located at 1220, 2202 ,
how far is it from the center fielder to third
base?
67. Distance between Moving Objects A Ford Focus and a
Freightliner truck leave an intersection at the same time.
The Focus heads east at an average speed of 30 miles per
hour, while the truck heads south at an average speed of
40 miles per hour. Find an expression for their distance apart
d (in miles) at the end of t hours.
68. Distance of a Moving Object from a Fixed Point A hot-air
balloon, headed due east at an average speed of 15 miles
per hour and at a constant altitude of 100 feet, passes over
an intersection (see the figure). Find an expression for
the distance d (measured in feet) from the balloon to the
intersection t seconds later.
62. P1 = 17, 22; P2 = 1 - 4, 02; P3 = 14, 62
East
63. Baseball A major league baseball “diamond” is actually a
square 90 feet on a side (see the figure). What is the distance
directly from home plate to second base (the diagonal of the
square)?
15 mph
100 ft
2nd base
90 ft
3rd base
Pitching
rubber
1st base
90 ft
Home plate
64. Little League Baseball The layout of a Little League
playing field is a square 60 feet on a side. How far is it
directly from home plate to second base (the diagonal of the
square)?
Source: Little League Baseball, Official Regulations and
Playing Rules, 2014.
65. Baseball Refer to Problem 63. Overlay a rectangular
coordinate system on a major league baseball diamond so
that the origin is at home plate, the positive x-axis lies in
the direction from home plate to first base, and the positive
y-axis lies in the direction from home plate to third
base.
(a) What are the coordinates of first base, second base, and
third base? Use feet as the unit of measurement.
(b) If the right fielder is located at 1310, 152, how far is it from
the right fielder to second base?
(c) If the center fielder is located at 1300, 3002,
how far is it from the center fielder to third
base?
69. Drafting Error When a draftsman draws three lines that
are to intersect at one point, the lines may not intersect as
intended and subsequently will form an error triangle. If this
error triangle is long and thin, one estimate for the location
of the desired point is the midpoint of the shortest side. The
figure shows one such error triangle.
y
(2.7, 1.7)
1.7
1.5
1.3
(2.6, 1.5)
(1.4, 1.3)
1.4
2.6 2.7
x
(a) Find an estimate for the desired intersection point.
(b) Find the length of the median for the midpoint found in
part (a). See Problem 55.
70. Net Sales The figure on page 157 illustrates how net sales
of Wal-Mart Stores, Inc., grew from 2007 through 2013. Use
the midpoint formula to estimate the net sales of Wal-Mart
Stores, Inc., in 2010. How does your result compare to the
reported value of $405 billion?
Source: Wal-Mart Stores, Inc., 2013 Annual Report
SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
71. Poverty Threshold Poverty thresholds are determined by
the U.S. Census Bureau. A poverty threshold represents the
minimum annual household income for a family not to be
considered poor. In 2003, the poverty threshold for a family of
four with two children under the age of 18 years was $18,660.
In 2013, the poverty threshold for a family of four with two
children under the age of 18 years was $23,624. Assuming
that poverty thresholds increase in a straight-line fashion, use
the midpoint formula to estimate the poverty threshold for a
family of four with two children under the age of 18 in
2008. How does your result compare to the actual poverty
threshold in 2008 of $21,834?
Source: U.S. Census Bureau
Net sales ($ billions)
Wal-Mart Stores, Inc.
Net sales ($ billions)
500
450
400
350
300 345
250
200
150
100
50
0
2007
466
2008
2009
2010
2011
2012
157
2013
Year
Explaining Concepts: Discussion and Writing
72. Write a paragraph that describes a Cartesian plane. Then
write a second paragraph that describes how to plot points
in the Cartesian plane. Your paragraphs should include
the terms “coordinate axes,” “ordered pair,” “coordinates,”
“plot,” “x-coordinate,” and “y-coordinate.”
Retain Your Knowledge
Problems 73–76 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
73. Determine the domain of the variable x in the expression
3x + 1
.
2x - 5
74. Find the real solution(s), if any, of the equation
3x2 - 7x - 20 = 0.
75. Multiply (7 + 3i)(1 - 2i). Write the answer in the form
a + bi.
76. Solve the inequality 5(x - 3) + 2x Ú 6(2x - 3) - 7.
Express the solution using interval notation. Graph the
solution set.
‘Are You Prepared?’ Answers
1. 0
2. 8
3. 5
4. 112 + 602 = 121 + 3600 = 3721 = 612
5. A =
1
bh
2
6. True
2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
PREPARING FOR THIS SECTION Before getting started, review the following:
r Solving Linear Equations (Section 1.1, pp. 82–85)
r Solve a Quadratic Equation by Factoring (Section 1.2,
pp. 93–94)
Now Work the ‘Are You Prepared?’ problems on page 164.
OBJECTIVES 1
2
3
4
Graph Equations by Plotting Points (p. 157)
Find Intercepts from a Graph (p. 159)
Find Intercepts from an Equation (p. 160)
Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis,
and the Origin (p. 160)
5 Know How to Graph Key Equations (p. 163)
1 Graph Equations by Plotting Points
An equation in two variables, say x and y, is a statement in which two expressions
involving x and y are equal. The expressions are called the sides of the equation. Since an
equation is a statement, it may be true or false, depending on the value of the variables.
Any values of x and y that result in a true statement are said to satisfy the equation.
For example, the following are all equations in two variables x and y:
x2 + y 2 = 5
2x - y = 6
y = 2x + 5
x2 = y
158
CHAPTER 2 Graphs
The first of these, x2 + y2 = 5, is satisfied for x = 1, y = 2, since 12 + 22 = 5.
Other choices of x and y, such as x = - 1, y = - 2, also satisfy this equation. It is
not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 ≠ 5.
The graph of an equation in two variables x and y consists of the set of points
in the xy-plane whose coordinates 1x, y2 satisfy the equation.
Graphs play an important role in helping us to visualize the relationships that
exist between two variables or quantities. Figure 11 shows the relation between
the level of risk in a stock portfolio and the average annual rate of return. From
the graph, we can see that when 30% of a portfolio of stocks is invested in foreign
companies, risk is minimized.
18.5
Average Annual Returns (%)
18
70%
17.5
(100% foreign)
50%
40%
16
15.5
100%
60%
17
16.5
90%
80%
30% (30% foreign/70% U.S.)
20%
15
14.5
10%
0% (100% U.S.)
14
Figure 11
13.5
13.5 14 14.5 15 15.5 16 16.5 17 17.5 18 18.5 19 19.5 20
Level of Risk (%)
Source: T. Rowe Price
EX AMPLE 1
Determining Whether a Point Is on the Graph of an Equation
Determine if the following points are on the graph of the equation 2x - y = 6.
(a) 12, 32
(b) 12, - 22
Solution
(a) For the point 12, 32 , check to see whether x = 2, y = 3 satisfies the equation
2x - y = 6.
2x - y = 2122 - 3 = 4 - 3 = 1 ≠ 6
The equation is not satisfied, so the point 12, 32 is not on the graph of 2x - y = 6.
(b) For the point 12, - 22 ,
2x - y = 2122 - 1 - 22 = 4 + 2 = 6
The equation is satisfied, so the point 12, - 22 is on the graph of 2x - y = 6.
Now Work
EX AMPLE 2
PROBLEM
13
r
Graphing an Equation by Plotting Points
Graph the equation: y = 2x + 5
Solution
y
25
(10, 25)
The graph consists of all points 1x, y2 that satisfy the equation. To locate some of
these points (and get an idea of the pattern of the graph), assign some numbers to x,
and find corresponding values for y.
If
(0, 5)
(1, 7)
– 25 (–5, – 5)
– 25
Figure 12 y = 2x + 5
25 x
Then
x = 0
y = 2102 + 5 = 5
x = 1
y = 2112 + 5 = 7
x = -5
y = 21 - 52 + 5 = - 5
x = 10
y = 21102 + 5 = 25
Point on Graph
10, 52
11, 72
1 - 5, - 52
110, 252
By plotting these points and then connecting them, we obtain the graph of the
equation (a line), as shown in Figure 12.
r
SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
159
Graphing an Equation by Plotting Points
EXAMPL E 3
Graph the equation: y = x2
Solution
Table 1
COMMENT Another way to obtain
the graph of an equation is to use a
graphing utility. Read Section 2, Using a
Graphing Utility to Graph Equations, in the
Appendix.
■
Graph
crosses
y-axis
y
Graph
crosses
x-axis
Table 1 provides several points on the graph. Plotting these points and connecting
them with a smooth curve gives the graph (a parabola) shown in Figure 13.
x
y = x2
(x, y)
-4
16
( - 4, 16)
-3
9
( - 3, 9)
-2
4
( - 2, 4)
-1
1
( - 1, 1)
0
0
(0, 0)
1
1
(1, 1)
2
4
(2, 4)
3
9
(3, 9)
4
16
(4, 16)
y
20
(– 4, 16)
(4, 16)
15
(–3, 9)
10
(3, 9)
5
(–2, 4)
(2, 4)
(1, 1)
(–1, 1)
(0, 0)
–4
4
Figure 13 y = x
2
x
r
The graphs of the equations shown in Figures 12 and 13 do not show all points.
For example, in Figure 12, the point 120, 452 is a part of the graph of y = 2x + 5,
but it is not shown. Since the graph of y = 2x + 5 can be extended out
indefinitely, we use arrows to indicate that the pattern shown continues. It is important
when illustrating a graph to present enough of the graph so that any viewer of the
illustration will “see” the rest of it as an obvious continuation of what is actually
there. This is referred to as a complete graph.
One way to obtain the complete graph of an equation is to plot enough points
on the graph for a pattern to become evident. Then these points are connected with
a smooth curve following the suggested pattern. But how many points are sufficient? Sometimes knowledge about the equation tells us. For example, we will learn
in the next section that if an equation is of the form y = mx + b, then its graph is a
line. In this case, only two points are needed to obtain the graph.
One purpose of this book is to investigate the properties of equations in order
to decide whether a graph is complete. Sometimes we shall graph equations by plotting
points. Shortly, we shall investigate various techniques that will enable us to graph
an equation without plotting so many points.
Two techniques that sometimes reduce the number of points required to graph
an equation involve finding intercepts and checking for symmetry.
2 Find Intercepts from a Graph
x
Graph
touches
x-axis
Intercepts
The points, if any, at which a graph crosses or touches the coordinate axes are called
the intercepts. See Figure 14. The x-coordinate of a point at which the graph crosses
or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the
graph crosses or touches the y-axis is a y-intercept.
Figure 14
EXAMPL E 4
Find the intercepts of the graph in Figure 15. What are its x-intercepts? What are its
y-intercepts?
y
4
Finding Intercepts from a Graph
(0, 3)
Solution The intercepts of the graph are the points
( 3–2 , 0)
4 (3, 0)
(0, 3.5)
Figure 15
1 - 3, 02,
(4.5, 0)
5 x
(0, 4–3 )
10, 32,
3
a , 0b ,
2
4
a0, - b ,
3
10, - 3.52,
14.5, 02
4
3
The x-intercepts are - 3, , and 4.5; the y-intercepts are - 3.5, - , and 3.
2
3
r
In Example 4, note the following usage: If the type of intercept (x- versus y-)
is not specified, then report the intercept as an ordered pair. However, if the type
of intercept is specified, then report the coordinate of the specified intercept. For
160
CHAPTER 2 Graphs
x-intercepts, report the x-coordinate of the intercept; for y-intercepts, report the
y-coordinate of the intercept.
Now Work
41(a)
PROBLEM
3 Find Intercepts from an Equation
The intercepts of a graph can be found from its equation by using the fact that
points on the x-axis have y-coordinates equal to 0, and points on the y-axis have
x-coordinates equal to 0.
COMMENT For many equations, finding
intercepts may not be so easy. In such
cases, a graphing utility can be used.
Read the first part of Section 3, Using
a Graphing Utility to Locate Intercepts and
Check for Symmetry, in the Appendix, to
find out how to locate intercepts using a
graphing utility.
■
EX AMPLE 5
Procedure for Finding Intercepts
1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in
the equation and solve for x, where x is a real number.
2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in
the equation and solve for y, where y is a real number.
Finding Intercepts from an Equation
Find the x-intercept(s) and the y-intercept(s) of the graph of y = x2 - 4. Then graph
y = x2 - 4 by plotting points.
Solution
To find the x-intercept(s), let y = 0 and obtain the equation
x2 - 4
1x + 22 1x - 2)
x + 2 = 0
or x - 2
x = -2
or
x
=
=
=
=
0
0
0
2
y = x 2 - 4 with y = 0
Factor.
Zero@Product Property
Solve.
The equation has two solutions, - 2 and 2. The x-intercepts are - 2 and 2.
To find the y-intercept(s), let x = 0 in the equation.
y = x2 - 4
= 02 - 4 = - 4
The y-intercept is - 4.
Since x2 Ú 0 for all x, we deduce from the equation y = x2 - 4 that y Ú - 4 for
all x. This information, the intercepts, and the points from Table 2 enable us to graph
y = x2 - 4. See Figure 16.
Table 2
x
y = x2 − 4
(x, y)
(–3, 5)
-3
5
-1
-3
( - 1, - 3)
1
-3
(1, - 3)
3
5
y
5
(3, 5)
( - 3, 5)
(2, 0)
(– 2, 0)
5 x
–5
(3, 5)
(– 1, – 3)
(1, – 3)
(0, – 4)
–5
Figure 16 y = x2 - 4
Now Work
PROBLEM
23
r
4 Test an Equation for Symmetry with Respect
to the x-Axis, the y-Axis, and the Origin
Another helpful tool for graphing equations by hand involves symmetry,
particularly symmetry with respect to the x-axis, the y-axis, and the origin.
Symmetry often occurs in nature. Consider the picture of the butterfly. Do you
see the symmetry?
SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
DEFINITION
161
A graph is said to be symmetric with respect to the x-axis if, for every point
1x, y2 on the graph, the point 1x, - y2 is also on the graph.
Figure 17 illustrates the definition. Note that when a graph is symmetric with
respect to the x-axis, the part of the graph above the x-axis is a reflection (or mirror
image) of the part below it, and vice versa.
y
(x, y)
(x, y )
Figure 17
Symmetry with respect to the x-axis
(x, –y)
(x, y )
x
(x, –y )
(x, –y )
Points Symmetric with Respect to the x-Axis
EXAMPL E 6
If a graph is symmetric with respect to the x-axis, and the point 13, 22 is on the
graph, then the point 13, - 22 is also on the graph.
r
DEFINITION
y
(–x, y)
A graph is said to be symmetric with respect to the y-axis if, for every point
1x, y2 on the graph, the point 1 - x, y2 is also on the graph.
(x, y )
(–x, y)
x
(x, y )
Figure 18 illustrates the definition. When a graph is symmetric with respect to
the y-axis, the part of the graph to the right of the y-axis is a reflection of the part
to the left of it, and vice versa.
Figure 18
Symmetry with respect to the y-axis
Points Symmetric with Respect to the y-Axis
EXAMPL E 7
If a graph is symmetric with respect to the y-axis and the point 15, 82 is on the graph,
then the point 1 - 5, 82 is also on the graph.
r
DEFINITION
y
A graph is said to be symmetric with respect to the origin if, for every point
1x, y2 on the graph, the point 1 - x, - y2 is also on the graph.
(x, y )
Figure 19 illustrates the definition. Symmetry with respect to the origin may be
viewed in three ways:
(x, y )
x
(–x, –y)
(–x, –y)
Figure 19
Symmetry with respect to the origin
EXAMPL E 8
1. As a reflection about the y-axis, followed by a reflection about the x-axis
2. As a projection along a line through the origin so that the distances from the
origin are equal
3. As half of a complete revolution about the origin
Points Symmetric with Respect to the Origin
If a graph is symmetric with respect to the origin, and the point 14, 22 is on the
graph, then the point 1 - 4, - 22 is also on the graph.
Now Work
PROBLEMS
31
AND
41(b)
r
When the graph of an equation is symmetric with respect to a coordinate axis
or the origin, the number of points that you need to plot in order to see the pattern
is reduced. For example, if the graph of an equation is symmetric with respect to the
162
CHAPTER 2 Graphs
y-axis, then once points to the right of the y-axis are plotted, an equal number of
points on the graph can be obtained by reflecting them about the y-axis. Because
of this, before we graph an equation, we should first determine whether it has any
symmetry. The following tests are used for this purpose.
Tests for Symmetry
To test the graph of an equation for symmetry with respect to the
x-Axis
y-Axis
Origin
EX AMPLE 9
Testing an Equation for Symmetry
Test y =
Solution
Replace y by - y in the equation and simplify. If an equivalent equation
results, the graph of the equation is symmetric with respect to the x-axis.
Replace x by - x in the equation and simplify. If an equivalent equation
results, the graph of the equation is symmetric with respect to the y-axis.
Replace x by - x and y by −y in the equation and simplify. If an
equivalent equation results, the graph of the equation is symmetric
with respect to the origin.
4x2
for symmetry.
x2 + 1
x-Axis: To test for symmetry with respect to the x-axis, replace y by - y. Since
4x2
4x2
-y = 2
is not equivalent to y = 2
, the graph of the equation is
x + 1
x + 1
not symmetric with respect to the x-axis.
y-Axis: To test for symmetry with respect to the y-axis, replace x by - x. Since
41 - x2 2
4x2
4x2
, the graph of the
y =
= 2
is equivalent to y = 2
2
1 - x2 + 1
x + 1
x + 1
equation is symmetric with respect to the y-axis.
Origin: To test for symmetry with respect to the origin, replace x by - x and y by - y.
-y =
41 - x2 2
1 - x2 2 + 1
4x2
x2 + 1
4x2
y = - 2
x + 1
-y =
Replace x by - x and y by - y.
Simplify.
Multiply both sides by - 1.
Since the result is not equivalent to the original equation, the graph of the
4x2
equation y = 2
is not symmetric with respect to the origin.
x + 1
r
Seeing the Concept
4x2
Figure 20 shows the graph of y = 2
using a graphing utility. Do you see the symmetry with
x + 1
respect to the y-axis?
5
25
Figure 20 y =
4x2
x + 1
5
25
2
Now Work
PROBLEM
61
SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
163
5 Know How to Graph Key Equations
The next three examples use intercepts, symmetry, and point plotting to obtain the
graphs of key equations. It is important to know the graphs of these key equations
because we use them later. The first of these is y = x3.
EX AM PL E 1 0
Graphing the Equation y = x 3 by Finding Intercepts, Checking
for Symmetry, and Plotting Points
Graph the equation y = x3 by plotting points. Find any intercepts and check for
symmetry first.
Solution
First, find the intercepts. When x = 0, then y = 0; and when y = 0, then x = 0. The
origin 10, 02 is the only intercept. Now test for symmetry.
Replace y by - y. Since - y = x3 is not equivalent to y = x3, the graph
is not symmetric with respect to the x-axis.
Replace x by - x. Since y = 1 - x2 3 = - x3 is not equivalent to y = x3,
the graph is not symmetric with respect to the y-axis.
Replace x by - x and y by - y. Since - y = 1 - x2 3 = - x3 is equivalent
to y = x3 (multiply both sides by - 1), the graph is symmetric with
respect to the origin.
x-Axis:
y-Axis:
y
8
(0, 0)
–6
(2, 8)
Origin:
(1, 1)
6
(– 1, – 1)
x
To graph y = x3, use the equation to obtain several points on the graph.
Because of the symmetry, we need to locate only points on the graph for which x Ú 0.
See Table 3. Since 11, 12 is on the graph, and the graph is symmetric with respect to
the origin, the point 1 - 1, - 12 is also on the graph. Plot the points from Table 3 and
use the symmetry. Figure 21 shows the graph.
Table 3
(– 2, – 8)
–8
Figure 21 y = x3
EX AM PL E 11
x
y = x3
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
8
(2, 8)
3
27
(3, 27)
Graphing the Equation x = y
r
2
(a) Graph the equation x = y2. Find any intercepts and check for symmetry first.
(b) Graph x = y2, y Ú 0.
Solution
(a) The lone intercept is 10, 02 . The graph is symmetric with respect to the x-axis.
(Do you see why? Replace y by - y.) Figure 22 shows the graph.
(b) If we restrict y so that y Ú 0, the equation x = y2, y Ú 0, may be written
equivalently as y = 1x. The portion of the graph of x = y2 in quadrant I is
therefore the graph of y = 1x. See Figure 23.
y
6
y
6
(9, 3)
(1, 1)
(0, 0)
6
Y1 5 √x
22
–2
(1, – 1)
10
26
Figure 24
Y2 5 2√x
(4, 2)
5
(1, 1)
10 x
(4, 2)
(9, 3)
(0, 0)
(4, – 2)
(9, – 3)
Figure 22 x = y2
–2
5
Figure 23 y = 1x
10 x
r
COMMENT To see the graph of the equation x = y 2 on a graphing calculator, you will need to graph two
equations: Y1 = 1x and Y2 = - 1x. We discuss why in Chapter 3. See Figure 24.
■
164
CHAPTER 2 Graphs
EX A MPL E 12
Graphing the Equation y =
1
x
1
. First, find any intercepts and check for symmetry.
x
Check for intercepts first. If we let x = 0, we obtain 0 in the denominator, which
makes y undefined. We conclude that there is no y-intercept. If we let y = 0, we get
1
the equation = 0, which has no solution. We conclude that there is no x-intercept.
x
1
The graph of y = does not cross or touch the coordinate axes.
x
Next check for symmetry:
1
1
x-Axis: Replacing y by - y yields - y = , which is not equivalent to y = .
x
x
1
1
y-Axis: Replacing x by - x yields y =
= - , which is not equivalent to
-x
x
1
y = .
x
1
Origin: Replacing x by - x and y by - y yields - y = - , which is equivalent to
x
1
y = . The graph is symmetric with respect to the origin.
x
Graph the equation y =
Solution
Table 4
1
x
x
y =
1
10
10
1
3
3
1
a , 3b
3
1
2
2
1
a , 2b
2
1
1
(1, 1)
2
1
2
1
a2, b
2
3
1
3
1
a3, b
3
10
1
10
(x, y)
a
1
, 10b
10
a10,
1
b
10
y
3
(––12 , 2)
(1, 1)
–3
(2, ––12 )
3
(–2, – ––12 )
(–1, –1)
x
Now set up Table 4, listing several points on the graph. Because of the
symmetry with respect to the origin, we use only positive values of x. From Table 4 we
1
infer that if x is a large and positive number, then y = is a positive number close
x
1
to 0. We also infer that if x is a positive number close to 0, then y = is a large and
x
positive number. Armed with this information, we can graph the equation.
1
Figure 25 illustrates some of these points and the graph of y = . Observe how
x
the absence of intercepts and the existence of symmetry with respect to the origin
were utilized.
r
1
COMMENT Refer to Example 2 in the Appendix, Section 3, for the graph of y =
found
x
using a graphing utility.
■
(– ––12 , –2)
–3
Figure 25 y =
1
x
2.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Solve the equation 21x + 32 - 1 = - 7. (pp. 82–85)
2. Solve the equation x2 - 9 = 0. (pp. 93–94)
Concepts and Vocabulary
3. The points, if any, at which a graph crosses or touches the
coordinate axes are called
.
8. True or False To find the y-intercepts of the graph of an
equation, let x = 0 and solve for y.
4. The x-intercepts of the graph of an equation are those
.
x-values for which
9. True or False The y-coordinate of a point at which the
graph crosses or touches the x-axis is an x-intercept.
5. If for every point 1x, y2 on the graph of an equation the
point 1 - x, y2 is also on the graph, then the graph is
symmetric with respect to the
.
6. If the graph of an equation is symmetric with respect to the
is also
y-axis and - 4 is an x-intercept of this graph, then
an x-intercept.
7. If the graph of an equation is symmetric with respect to the
origin and 13, - 42 is a point on the graph, then
is
also a point on the graph.
10. True or False If a graph is symmetric with respect to the
x-axis, then it cannot be symmetric with respect to the
y-axis.
11. Given that the intercepts of a graph are (–4, 0) and (0, 5),
choose the statement that is true.
(a) The y-intercept is –4, and the x-intercept is 5.
(b) The y-intercepts are –4 and 5.
(c) The x-intercepts are –4 and 5.
(d) The x-intercept is –4, and the y-intercept is 5.
SECTION 2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
12. To test whether the graph of an equation is symmetric with
respect to the origin, replace
in the equation and
simplify. If an equivalent equation results, then the graph is
symmetric with respect to the origin.
(a)
(b)
(c)
(d)
165
x by –x
y by –y
x by –x and y by –y
x by –y and y by –x
Skill Building
In Problems 13–18, determine which of the given points are on the graph of the equation.
13. Equation: y = x4 - 1x
Points: 10, 02; 11, 12; 12, 42
14. Equation: y = x3 - 21x
Points: 10, 02; 11, 12; 11, - 12
15. Equation: y2 = x2 + 9
Points: 10, 32; 13, 02; 1 - 3, 02
16. Equation: y3 = x + 1
Points: 11, 22; 10, 12; 1 - 1, 02
17. Equation: x2 + y2 = 4
Points: 10, 22; 1 - 2, 22;
18. Equation: x2 + 4y2 = 4
1 22, 22 2
1
Points: 10, 12; 12, 02; a2, b
2
In Problems 19–30, find the intercepts and graph each equation by plotting points. Be sure to label the intercepts.
19. y = x + 2
20. y = x - 6
21. y = 2x + 8
22. y = 3x - 9
23. y = x - 1
24. y = x - 9
25. y = - x + 4
26. y = - x2 + 1
27. 2x + 3y = 6
28. 5x + 2y = 10
29. 9x2 + 4y = 36
30. 4x2 + y = 4
2
2
2
In Problems 31–40, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin.
31. 13, 42
32. 15, 32
36. 1 - 1, - 12
33. 1 - 2, 12
37. 1 - 3, - 42
34. 14, - 22
38. 14, 02
35. 15, - 22
39. 10, - 32
40. 1 - 3, 02
In Problems 41–52, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect to
the x-axis, the y-axis, or the origin.
41.
42.
y
3
46.
y
3
3x
–3
––
3x
2
1
x
––
2
3
47.
50.
3x
3
3
y
40
x
3x
3
3x
–3
48.
y
6
6
6
6
–3
y
3
3 x
4
y
3
–3
49.
y
4
3
–3
45.
44.
y
1
3
3x
–3
43.
y
3
40
51.
y
3
52.
8
−2
3x
3
4
3
2
−4
4
−4
−8
In Problems 53–56, draw a complete graph so that it has the type of symmetry indicated.
53. y-axis
54. x-axis
y
9
y
55. Origin y
5
4
(0, 2)
(5, 3)
(0, 0)
(–4, 0)
5x
–5
9x
(2, –5)
–9
–9
(0, –9)
56. y-axis
, 2)
(––
2
y
(0, 4)
4
(2, 2)
(, 0)
x
(0, 0)
3x
–3
–5
–4
–2
166
CHAPTER 2 Graphs
In Problems 57–72, list the intercepts and test for symmetry.
57. y2 = x + 4
58. y2 = x + 9
3
59. y = 2
x
5
60. y = 2
x
61. x2 + y - 9 = 0
62. x2 - y - 4 = 0
63. 9x2 + 4y2 = 36
64. 4x2 + y2 = 4
65. y = x3 - 27
66. y = x4 - 1
67. y = x2 - 3x - 4
68. y = x2 + 4
69. y =
3x
x + 9
2
70. y =
x2 - 4
2x
71. y =
- x3
x - 9
2
72. y =
x4 + 1
2x5
In Problems 73–76, draw a quick sketch of the graph of each equation.
73. y = x3
74. x = y2
77. If 1a, 42 is a point on the graph of y = x2 + 3x, what is a?
75. y = 1x
76. y =
1
x
78. If 1a, - 52 is a point on the graph of y = x2 + 6x, what is a?
Applications and Extensions
79. Given that the point (1, 2) is on the graph of an equation
that is symmetric with respect to the origin, what other point
is on the graph?
80. If the graph of an equation is symmetric with respect to the
y-axis and 6 is an x-intercept of this graph, name another
x-intercept.
81. If the graph of an equation is symmetric with respect to the
origin and - 4 is an x-intercept of this graph, name another
x-intercept.
82. If the graph of an equation is symmetric with respect to the
x-axis and 2 is a y-intercept, name another y-intercept.
83. Microphones In studios and on stages, cardioid microphones
are often preferred for the richness they add to voices and
for their ability to reduce the level of sound from the sides
and rear of the microphone. Suppose one such cardioid
pattern is given by the equation 1x2 + y2 - x2 2 = x2 + y2.
(a) Find the intercepts of the graph of the equation.
(b) Test for symmetry with respect to the x-axis, the y-axis,
and the origin.
Source: www.notaviva.com
84. Solar Energy The solar electric generating systems at
Kramer Junction, California, use parabolic troughs to heat
a heat-transfer fluid to a high temperature. This fluid is used
to generate steam that drives a power conversion system to
produce electricity. For troughs 7.5 feet wide, an equation
for the cross section is 16y2 = 120x - 225.
(a) Find the intercepts of the graph of the equation.
(b) Test for symmetry with respect to the x-axis, the y-axis,
and the origin.
Source: U.S. Department of Energy
Explaining Concepts: Discussion and Writing
85. (a) Graph y = 2x2 , y = x, y = 0 x 0 , and y = 1 1x2 2, noting
which graphs are the same.
(b) Explain why the graphs of y = 2x2 and y = 0 x 0 are the
same.
(c) Explain why the graphs of y = x and y = 1 1x2 2 are
not the same.
(d) Explain why the graphs of y = 2x2 and y = x are not
the same.
86. Explain what is meant by a complete graph.
87. Draw a graph of an equation that contains two x-intercepts;
at one the graph crosses the x-axis, and at the other the
graph touches the x-axis.
88. Make up an equation with the intercepts 12, 02, 14, 02 , and
10, 12 . Compare your equation with a friend’s equation.
Comment on any similarities.
89. Draw a graph that contains the points 1 - 2, - 12, 10, 12,
11, 32 , and 13, 52 . Compare your graph with those of other
students. Are most of the graphs almost straight lines? How
many are “curved”? Discuss the various ways in which these
points might be connected.
90. An equation is being tested for symmetry with respect to
the x-axis, the y-axis, and the origin. Explain why, if two of
these symmetries are present, the remaining one must also
be present.
91. Draw a graph that contains the points ( - 2, 5), ( - 1, 3),
and (0, 2) and is symmetric with respect to the y-axis.
Compare your graph with those of other students; comment
on any similarities. Can a graph contain these points and be
symmetric with respect to the x-axis? the origin? Why or
why not?
SECTION 2.3 Lines
167
Retain Your Knowledge
Problems 92–95 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
x + y
94. Simplify: 2- 196
if x = 6 and y = - 2.
92. Find the value of
x - y
95. Solve x2 - 8x + 4 = 0 by completing the square.
2
93. Factor 3x - 30x + 75 completely.
‘Are You Prepared?’ Answers
1. 5 - 66
2. 5 - 3, 36
2.3 Lines
OBJECTIVES 1
2
3
4
5
6
7
8
9
10
Calculate and Interpret the Slope of a Line (p. 167)
Graph Lines Given a Point and the Slope (p. 170)
Find the Equation of a Vertical Line (p. 170)
Use the Point–Slope Form of a Line; Identify Horizontal Lines (p. 171)
Find the Equation of a Line Given Two Points (p. 172)
Write the Equation of a Line in Slope–Intercept Form (p. 172)
Identify the Slope and y-Intercept of a Line from Its Equation (p. 173)
Graph Lines Written in General Form Using Intercepts (p. 174)
Find Equations of Parallel Lines (p. 175)
Find Equations of Perpendicular Lines (p. 176)
In this section we study a certain type of equation that contains two variables, called
a linear equation, and its graph, a line.
Line
Rise
Run
Figure 26
DEFINITION
1 Calculate and Interpret the Slope of a Line
Consider the staircase illustrated in Figure 26. Each step contains exactly the same
horizontal run and the same vertical rise. The ratio of the rise to the run, called the
slope, is a numerical measure of the steepness of the staircase. For example, if the
run is increased and the rise remains the same, the staircase becomes less steep. If
the run is kept the same but the rise is increased, the staircase becomes more steep.
This important characteristic of a line is best defined using rectangular coordinates.
Let P = 1x1 , y1 2 and Q = 1x2 , y2 2 be two distinct points. If x1 ≠ x2 , the
slope m of the nonvertical line L containing P and Q is defined by the formula
m =
y2 - y1
x2 - x1
x1 ≠ x2
(1)
If x1 = x2 , then L is a vertical line and the slope m of L is undefined (since this
results in division by 0).
Figure 27(a) on page 168 provides an illustration of the slope of a nonvertical
line; Figure 27(b) illustrates a vertical line.
168
CHAPTER 2 Graphs
L
L
Q = (x 2, y2)
y2
Run = x2 – x1
x1
Figure 27
Q = (x 1, y2)
y1
P = (x 1, y1)
Rise = y2 – y1
P = (x 1, y1)
y1
y2
x2
(a) Slope of L is m =
x1
y2 – y1
_______ , x Z x
x2 – x1 1 2
(b) Slope is undefined; L is vertical
As Figure 27(a) illustrates, the slope m of a nonvertical line may be viewed as
In Words
The symbol ∆ is the Greek uppercase
letter delta. In mathematics, ∆
∆y
is read “change in,” so
is read
∆x
“change in y divided by change in x.”
m =
y2 - y1
Rise
=
x2 - x1
Run
or as m =
Change in y
y2 - y1
∆y
=
=
x2 - x1
Change in x
∆x
That is, the slope m of a nonvertical line measures the amount y changes when x
∆y
is called the average rate of change of y
changes from x1 to x2. The expression
∆x
with respect to x.
Two comments about computing the slope of a nonvertical line may prove
helpful:
1. Any two distinct points on the line can be used to compute the slope of the
line. (See Figure 28 for justification.) Since any two distinct points can be used
to compute the slope of a line, the average rate of change of a line is always the
same number.
y
Figure 28
Triangles ABC and PQR are similar (equal
angles), so ratios of corresponding sides
are proportional. Then
y2 - y1
Slope using P and Q =
=
x2 - x1
d(B, C)
= Slope using A and B
d(A, C)
Q = (x 2, y2)
y2 – y1
P = (x 1, y1)
B
x2 – x1
R
x
A
C
2. The slope of a line may be computed from P = 1x1 , y1 2 to Q = 1x2 , y2 2 or
from Q to P because
y2 - y1
y1 - y2
=
x2 - x1
x1 - x2
EX AMPLE 1
Finding and Interpreting the Slope of a Line Given Two Points
The slope m of the line containing the points 11, 22 and 15, - 32 may be computed as
m =
-3 - 2
-5
5
=
= 5 - 1
4
4
or as m =
2 - 1 - 32
5
5
=
= 1 - 5
-4
4
For every 4-unit change in x, y will change by - 5 units. That is, if x increases by
4 units, then y will decrease by 5 units. The average rate of change of y with respect
5
to x is - .
4
r
Now Work
PROBLEMS
13
AND
19
SECTION 2.3 Lines
169
Finding the Slopes of Various Lines Containing the Same Point (2, 3)
EXAMPL E 2
Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of
points. Graph all four lines on the same set of coordinate axes.
L1 :
L2 :
L3 :
L4 :
Solution
P
P
P
P
=
=
=
=
12, 32
12, 32
12, 32
12, 32
Q1
Q2
Q3
Q4
=
=
=
=
1 - 1, - 22
13, - 12
15, 32
12, 52
Let m1 , m2 , m3 , and m4 denote the slopes of the lines L1 , L2, L3 , and L4 ,
respectively. Then
-2 - 3
-5
5
A rise of 5 divided by a run of 3
=
=
-1 - 2
-3
3
-1 - 3
-4
m2 =
=
= -4
3 - 2
1
0
3 - 3
= = 0
m3 =
5 - 2
3
m4 is undefined because x1 = x2 = 2
m1 =
L2
L4
L1
y
5
Q4 5 (2, 5)
P 5 (2, 3)
m3 5 0
L3
Q3 5 (5, 3)
25
m1 5
5
3
The graphs of these lines are given in Figure 29.
5 x
Q2 5 (3, 21)
Q1 5 (21, 22)
23
m4 undefined
r
Figure 29 illustrates the following facts:
m2 5 24
1. When the slope of a line is positive, the line slants upward from left to right
1L1 2 .
2. When the slope of a line is negative, the line slants downward from left to
right 1L2 2 .
3. When the slope is 0, the line is horizontal 1L3 2 .
4. When the slope is undefined, the line is vertical 1L4 2 .
Figure 29
Seeing the Concept
Y6 5 6x
Y5 5 2x Y4 5 x
Y3 5
2
On the same screen, graph the following equations:
Y1 = 0
1
Y2 = x
4
1
Y3 = x
2
Y4 = x
Y5 = 2x
Y6 = 6x
1
x
2
1
Y2 5 4 x
23
3
Y1 5 0
22
Figure 30
Slope of line is 0.
1
Slope of line is .
4
1
Slope of line is .
2
Slope of line is 1.
Slope of line is 2.
Slope of line is 6.
See Figure 30.
Seeing the Concept
Y4 5 2x
Y6 5 26x
Y5 5 22x
On the same screen, graph the following equations:
Y1 = 0
1
Y3 5 2 2 x
2
1
Y2 5 2 4 x
23
Slope of line is -
Y3 =
Slope of line is -
Y4 =
Y5 =
Y6 =
22
Figure 31
Y2 = -
3
Y1 5 0
See Figure 31.
1
x
4
1
- x
2
-x
- 2x
- 6x
Slope of line is 0.
1
.
4
1
.
2
Slope of line is - 1.
Slope of line is - 2.
Slope of line is - 6.
170
CHAPTER 2 Graphs
Figures 30 and 31 on page 169 illustrate that the closer the line is to the vertical
position, the greater the magnitude of the slope.
2 Graph Lines Given a Point and the Slope
EX AMPLE 3
Graphing a Line Given a Point and a Slope
Draw a graph of the line that contains the point 13, 22 and has a slope of:
(a)
Solution
y
6
(7, 5)
Rise = 3
(3, 2)
Run = 4
(b) -
Rise
3
. The slope means that for every horizontal movement (run)
Run
4
of 4 units to the right, there will be a vertical movement (rise) of 3 units. Start at
the given point 13, 22 and move 4 units to the right and 3 units up, arriving at the
point 17, 52 . Drawing the line through this point and the point 13, 22 gives the
graph. See Figure 32.
(b) The fact that the slope is
-
y
(–2, 6)
6
Rise = 4
4
-4
Rise
=
=
5
5
Run
means that for every horizontal movement of 5 units to the right, there will be
a corresponding vertical movement of - 4 units (a downward movement). Start
at the given point 13, 22 and move 5 units to the right and then 4 units down,
arriving at the point 18, - 22 . Drawing the line through these points gives the
graph. See Figure 33.
Alternatively, consider that
Figure 32
(3, 2) Run = 5
Run = –5
-
Rise = –4
10 x
–2
–2
4
5
(a) Slope =
10 x
5
–2
3
4
4
4
Rise
=
=
5
-5
Run
so for every horizontal movement of - 5 units (a movement to the left), there
will be a corresponding vertical movement of 4 units (upward). This approach
leads to the point 1 - 2, 62 , which is also on the graph of the line in Figure 33.
(8, –2)
Figure 33
Now Work
PROBLEM
25 (graph the line)
r
3 Find the Equation of a Vertical Line
EX AMPLE 4
Graphing a Line
Graph the equation: x = 3
Solution
To graph x = 3, we find all points 1x, y2 in the plane for which x = 3. No
matter what y-coordinate is used, the corresponding x-coordinate always equals 3.
Consequently, the graph of the equation x = 3 is a vertical line with x-intercept 3
and undefined slope. See Figure 34.
y
4
(3, 3)
(3, 2)
(3, 1)
1
1
(3, 0)
(3, 1)
Figure 34 x = 3
5 x
r
SECTION 2.3 Lines
171
Example 4 suggests the following result:
THEOREM
Equation of a Vertical Line
A vertical line is given by an equation of the form
x = a
where a is the x-intercept.
COMMENT To graph an equation using a graphing utility, we need to express the equation in the form
y = 5 expression in x6 . But x = 3 cannot be put in this form. To overcome this, most graphing utilities
have special commands for drawing vertical lines. DRAW, LINE, PLOT, and VERT are among the more common
ones. Consult your manual to determine the correct methodology for your graphing utility.
■
y
L
(x, y )
Let L be a nonvertical line with slope m that contains the point 1x1 , y1 2. See Figure 35.
For any other point 1x, y2 on L, we have
y – y1
(x 1, y1)
4 Use the Point–Slope Form of a Line; Identify Horizontal Lines
x – x1
m =
x
y - y1
x - x1
or y - y1 = m1x - x1 2
Figure 35
THEOREM
Point–Slope Form of an Equation of a Line
An equation of a nonvertical line with slope m that contains the point 1x1 , y1 2 is
y - y1 = m1x - x1 2
EXAMPL E 5
y
6
(2, 6)
Using the Point–Slope Form of a Line
An equation of the line with slope 4 that contains the point 11, 22 can be found by
using the point–slope form with m = 4, x1 = 1, and y1 = 2.
y - y1 = m1x - x1 2
Rise 5 4
(1, 2)
y - 2 = 41x - 12
Run 5 1
22
y = 4x - 2
5
(2)
m = 4, x1 = 1, y1 = 2
Solve for y.
x
See Figure 36 for the graph.
Now Work
PROBLEM
25 (find the point-slope form)
r
Figure 36 y = 4x - 2
EXAMPL E 6
Solution
y
(3, 2)
1
3
Figure 37 y = 2
Find an equation of the horizontal line containing the point 13, 22 .
Because all the y-values are equal on a horizontal line, the slope of a horizontal line
is 0. To get an equation, use the point–slope form with m = 0, x1 = 3, and y1 = 2.
y - y1 = m1x - x1 2
4
–1
Finding the Equation of a Horizontal Line
5 x
y - 2 = 0 # 1x - 32 m = 0, x1 = 3, and y1 = 2
y - 2 = 0
y = 2
See Figure 37 for the graph.
r
172
CHAPTER 2 Graphs
Example 6 suggests the following result:
THEOREM
Equation of a Horizontal Line
A horizontal line is given by an equation of the form
y = b
where b is the y-intercept.
5 Find the Equation of a Line Given Two Points
EX A MPL E 7
Solution
Finding an Equation of a Line Given Two Points
Find an equation of the line containing the points 12, 32 and 1 - 4, 52 . Graph the
line.
First compute the slope of the line.
m =
2
1
5 - 3
=
= -4 - 2
-6
3
(–4, 5)
(2, 3)
2
–2
y2 - y1
x2 - x1
1
Use the point 12, 32 and the slope m = - to get the point–slope form of the
3
equation of the line.
y
–4
m =
10
x
1
Figure 38 y - 3 = - (x - 2)
3
y - 3 = -
1
1x - 22
3
y - y1 = m(x - x1)
See Figure 38 for the graph.
r
In the solution to Example 7, we could have used the other point, 1 - 4, 52 ,
instead of the point 12, 32 . The equation that results, although it looks different, is
equivalent to the equation that we obtained in the example. (Try it for yourself.)
Now Work
PROBLEM
39
6 Write the Equation of a Line in Slope–Intercept Form
Another useful equation of a line is obtained when the slope m and y-intercept b
are known. In this event, both the slope m of the line and a point 10, b2 on the line
are known; then use the point–slope form, equation (2), to obtain the following
equation:
y - b = m1x - 02
THEOREM
or y = mx + b
Slope–Intercept Form of an Equation of a Line
An equation of a line with slope m and y-intercept b is
y = mx + b
Now Work
PROBLEMS 47 AND 53 (express answer
in slope–intercept form)
(3)
SECTION 2.3 Lines
Y5 5 23x 1 2
Y4 5 3x 1 2
Seeing the Concept
Y2 5x 1 2
To see the role that the slope m plays, graph the following lines on the same screen.
4
Y3 52x 1 2
Y1 5 2
26
173
6
Y1 = 2
m = 0
Y2 = x + 2
m = 1
Y3 = - x + 2
m = -1
Y4 = 3x + 2
m = 3
Y5 = - 3x + 2 m = - 3
See Figure 39. What do you conclude about the lines y = mx + 2?
24
Figure 39 y = mx + 2
Y2 5 2x 1 1 Y1 5 2x Y3 5 2x 2 1
Seeing the Concept
To see the role of the y-intercept b, graph the following lines on the same screen.
Y1 = 2x
4
b = 0
Y2 = 2x + 1 b = 1
Y4 5 2x 1 4
Y3 = 2x - 1 b = - 1
26
Y4 = 2x + 4 b = 4
6
24
Y5 = 2x - 4 b = - 4
Y5 5 2x 2 4
See Figure 40. What do you conclude about the lines y = 2x + b?
Figure 40 y = 2x + b
7 Identify the Slope and y-Intercept of a Line from Its Equation
When the equation of a line is written in slope–intercept form, it is easy to find the
slope m and y-intercept b of the line. For example, suppose that the equation of a
line is
y = - 2x + 7
Compare this equation to y = mx + b.
y = - 2x + 7
y =
c
c
mx
+ b
The slope of this line is - 2 and its y-intercept is 7.
Now Work
EXAMPL E 8
PROBLEM
73
Finding the Slope and y-Intercept
Find the slope m and y-intercept b of the equation 2x + 4y = 8. Graph the equation.
Solution
To obtain the slope and y-intercept, write the equation in slope–intercept form by
solving for y.
2x + 4y = 8
4y = - 2x + 8
y = -
y
4
(0, 2)
–3
2
1
(2, 1)
3
1
Figure 41 y = - x + 2
2
x
1
x + 2
2
y = mx + b
1
The coefficient of x, - , is the slope, and the constant, 2, is the y-intercept. Graph
2
1
the line with y-intercept 2 and with slope - . Starting at the point 10, 22, go to the
2
right 2 units and then down 1 unit to the point 12, 12 . Draw the line through
these points. See Figure 41.
Now Work
PROBLEM
79
r
174
CHAPTER 2 Graphs
8 Graph Lines Written in General Form Using Intercepts
Refer to Example 8. The form of the equation of the line 2x + 4y = 8 is called the
general form.
DEFINITION
The equation of a line is in general form* when it is written as
Ax + By = C
(4)
where A, B, and C are real numbers and A and B are not both 0.
If B = 0 in equation (4), then A ≠ 0 and the graph of the equation is a vertical
C
line: x = . If B ≠ 0 in equation (4), then we can solve the equation for y and
A
write the equation in slope–intercept form as we did in Example 8.
Another approach to graphing equation (4) is to find its intercepts. Remember,
the intercepts of the graph of an equation are the points where the graph crosses or
touches a coordinate axis.
EX AMPLE 9
Graphing an Equation in General Form Using Its Intercepts
Graph the equation 2x + 4y = 8 by finding its intercepts.
Solution
To obtain the x-intercept, let y = 0 in the equation and solve for x.
2x + 4y = 8
2x + 4102 = 8
Let y = 0.
2x = 8
x = 4
Divide both sides by 2.
The x-intercept is 4, and the point 14, 02 is on the graph of the equation.
To obtain the y-intercept, let x = 0 in the equation and solve for y.
2x + 4y = 8
y
4
2102 + 4y = 8
4y = 8
(0, 2)
y = 2
(4, 0)
–3
Let x = 0.
3
Figure 42 2x + 4y = 8
x
Divide both sides by 4.
The y-intercept is 2, and the point 10, 22 is on the graph of the equation.
Plot the points 14, 02 and 10, 22 and draw the line through the points. See
Figure 42.
Now Work
PROBLEM
r
93
Every line has an equation that is equivalent to an equation written in general
form. For example, a vertical line whose equation is
x = a
can be written in the general form
1#x + 0#y = a
A = 1, B = 0, C = a
A horizontal line whose equation is
y = b
can be written in the general form
0#x + 1#y = b
*Some texts use the term standard form.
A = 0, B = 1, C = b
SECTION 2.3 Lines
175
Lines that are neither vertical nor horizontal have general equations of the form
Ax + By = C
A ≠ 0 and B ≠ 0
Because the equation of every line can be written in general form, any equation
equivalent to equation (4) is called a linear equation.
9 Find Equations of Parallel Lines
y
Rise
Rise
When two lines (in the plane) do not intersect (that is, they have no points in
common), they are parallel. Look at Figure 43. There we have drawn two parallel
lines and have constructed two right triangles by drawing sides parallel to the
coordinate axes. The right triangles are similar. (Do you see why? Two angles are
equal.) Because the triangles are similar, the ratios of corresponding sides are equal.
Run
Run
x
THEOREM Criteria for Parallel Lines
Two nonvertical lines are parallel if and only if their slopes are equal and they
have different y-intercepts.
Figure 43 Parallel lines
The use of the phrase “if and only if” in the preceding theorem means that
actually two statements are being made, one the converse of the other.
If two nonvertical lines are parallel, then their slopes are equal and they
have different y-intercepts.
If two nonvertical lines have equal slopes and they have different y-intercepts,
then they are parallel.
EX AM PL E 1 0
Showing That Two Lines Are Parallel
Show that the lines given by the following equations are parallel.
L1 : 2x + 3y = 6
Solution
To determine whether these lines have equal slopes and different y-intercepts, write
each equation in slope–intercept form.
L1 : 2x + 3y = 6
y
5
L2 : 4x + 6y = 0
3y = - 2x + 6
y = -
5
L2 : 4x + 6y = 0
5 x
L1
L2
Figure 44
Slope = -
2
;
3
6y = - 4x
2
x + 2
3
y@intercept = 2
y = Slope = -
2
;
3
2
x
3
y@intercept = 0
2
Because these lines have the same slope, - , but different y-intercepts, the lines are
3
parallel. See Figure 44.
r
EX AM PL E 1 1
Solution
Finding a Line That Is Parallel to a Given Line
Find an equation for the line that contains the point 12, - 32 and is parallel to the
line 2x + y = 6.
Since the two lines are to be parallel, the slope of the line being sought equals the
slope of the line 2x + y = 6. Begin by writing the equation of the line 2x + y = 6
in slope–intercept form.
2x + y = 6
y = - 2x + 6
176
CHAPTER 2 Graphs
The slope is - 2. Since the line being sought also has slope - 2 and contains the
point 12, - 32 , use the point–slope form to obtain its equation.
y
y - y1 = m1x - x1 2
6
y - 1 - 32 = - 21x - 22
y + 3 = - 2x + 4
6
y = - 2x + 1
6 x
2x y 6
(2, 3)
5
2x y 1
Figure 45
2x + y = 1
Point–slope form
m = - 2, x1 = 2, y1 = - 3
Simplify.
Slope–intercept form
General form
This line is parallel to the line 2x + y = 6 and contains the point 12, - 32 . See
Figure 45.
Now Work
PROBLEM
r
61
y
10 Find Equations of Perpendicular Lines
When two lines intersect at a right angle (90°), they are perpendicular. See
Figure 46.
The following result gives a condition, in terms of their slopes, for two lines to
be perpendicular.
90°
x
Figure 46 Perpendicular lines
THEOREM
Criterion for Perpendicular Lines
Two nonvertical lines are perpendicular if and only if the product of their
slopes is - 1.
Here we shall prove the “only if” part of the statement:
If two nonvertical lines are perpendicular, then the product of their slopes
is - 1.
In Problem 130 you are asked to prove the “if” part of the theorem:
If two nonvertical lines have slopes whose product is - 1, then the lines are
perpendicular.
y
Slope m2
A = (1, m2)
Slope m1
Rise = m 2
x
Run = 1
O
1
3 d 1O, A2 4 2 + 3 d 1O, B2 4 2 = 3 d 1A, B2 4 2
(5)
Rise = m1
B = (1, m1)
Figure 47
Proof Let m1 and m2 denote the slopes of the two lines. There is no loss in generality
(that is, neither the angle nor the slopes are affected) if we situate the lines so that
they meet at the origin. See Figure 47. The point A = 11, m2 2 is on the line having
slope m2 , and the point B = 11, m1 2 is on the line having slope m1 . (Do you see
why this must be true?)
Suppose that the lines are perpendicular. Then triangle OAB is a right triangle.
As a result of the Pythagorean Theorem, it follows that
Using the distance formula, the squares of these distances are
3 d 1O, A2 4 2 = 11 - 02 2 + 1m2 - 02 2 = 1 + m22
3 d 1O, B2 4 2 = 11 - 02 2 + 1m1 - 02 2 = 1 + m21
3 d 1A, B2 4 2 = 11 - 12 2 + 1m2 - m1 2 2 = m22 - 2m1 m2 + m21
SECTION 2.3 Lines
177
Using these facts in equation (5), we get
11
+ m22 2 +
11
+ m21 2 = m22 - 2m1 m2 + m21
which, upon simplification, can be written as
m1 m2 = - 1
If the lines are perpendicular, the product of their slopes is - 1.
■
You may find it easier to remember the condition for two nonvertical lines to be
perpendicular by observing that the equality m1 m2 = - 1 means that m1 and m2 are
1
1
negative reciprocals of each other; that is, m1 = and m2 = .
m2
m1
Finding the Slope of a Line Perpendicular to Another Line
EX AM PL E 1 2
3
2
If a line has slope , any line having slope - is perpendicular to it.
2
3
r
Finding the Equation of a Line Perpendicular to a Given Line
EX AM PL E 1 3
Find an equation of the line that contains the point 11, - 22 and is perpendicular to
the line x + 3y = 6. Graph the two lines.
Solution
First write the equation of the given line in slope–intercept form to find its slope.
x + 3y = 6
3y = - x + 6
y = -
1
x + 2
3
Proceed to solve for y.
Place in the form y = mx + b.
1
The given line has slope - . Any line perpendicular to this line will have slope 3.
3
Because the point 11, - 22 is on this line with slope 3, use the point–slope form of
the equation of a line.
y
y 3x 5
6
x 3y 6
y - y1 = m1x - x1 2
y - 1 - 22 = 31x - 12
4
2
2
2
4
6
To obtain other forms of the equation, proceed as follows:
(1, 2)
y + 2 = 3x - 3
y = 3x - 5
4
3x - y = 5
Figure 48
m = 3, x1 = 1, y1 = - 2
y + 2 = 31x - 12
x
2
Point–slope form
Figure 48 shows the graphs.
Now Work
PROBLEM
Simplify.
Slope–intercept form
General form
r
67
WARNING Be sure to use a square screen when you use a graphing calculator to graph perpendicular
lines. Otherwise, the angle between the two lines will appear distorted. A discussion of square screens
is given in Section 5 of the Appendix.
■
178
CHAPTER 2 Graphs
2.3 Assess Your Understanding
Concepts and Vocabulary
1. The slope of a vertical line is
horizontal line is
.
; the slope of a
2. For the line 2x + 3y = 6, the x-intercept is
y-intercept is
.
and the
3. True or False The equation 3x + 4y = 6 is written in
general form.
4. True or False The slope of the line 2y = 3x + 5 is 3.
5. True or False The point 11, 22 is on the line 2x + y = 4.
6. Two nonvertical lines have slopes m1 and m2 , respectively.
The lines are parallel if
and the
are unequal; the lines are perpendicular if
.
7. The lines y = 2x + 3 and y = ax + 5 are parallel if
.
a =
8. The lines y = 2x - 1 and y = ax + 2 are perpendicular if
a =
.
9. True or False Perpendicular lines have slopes that are
reciprocals of one another.
10. Choose the formula for finding the slope m of a nonvertical
line that contains the two distinct points (x1, y1) and (x2, y2).
y2 - x2
(a) m =
x1 ≠ y1
y1 - x1
y2 - x1
(b) m =
y1 ≠ x2
x2 - y1
x2 - x1
(c) m =
y1 ≠ y2
y2 - y1
y2 - y1
(d) m =
x1 ≠ x2
x2 - x1
11. If a line slants downward from left to right, then which of the
following describes its slope?
(a) positive
(b) zero
(c) negative
(d) undefined
12. Choose the correct statement about the graph of the line
y = - 3.
(a) The graph is vertical with x-intercept - 3.
(b) The graph is horizontal with y-intercept - 3.
(c) The graph is vertical with y-intercept - 3.
(d) The graph is horizontal with x-intercept - 3.
Skill Building
In Problems 13–16, (a) find the slope of the line and (b) interpret the slope.
13.
14.
y
2
(–2, 1) 2
(2, 1)
16.
y
(–2, 2)
2
(1, 1)
y
(–1, 1)
2
(2, 2)
(0, 0)
(0, 0)
–2
15.
y
2
–1
x
–2
x
2
–1
–2
2
–1
x
–2
x
2
–1
In Problems 17–24, plot each pair of points and determine the slope of the line containing them. Graph the line.
17. 12, 32; 14, 02
21. 1 - 3, - 12; 12, - 12
18. 14, 22; 13, 42
19. 1 - 2, 32; 12, 12
22. 14, 22; 1 - 5, 22
23. 1 - 1, 22; 1 - 1, - 22
20. 1 - 1, 12; 12, 32
24. 12, 02; 12, 22
In Problems 25–32, graph the line that contains the point P and has slope m. In Problems 25–30, find the point-slope form of the
equation of the line. In Problems 31 and 32, find an equation of the line.
25. P = 11, 22; m = 3
26. P = 12, 12; m = 4
29. P = 1 - 1, 32; m = 0
30. P = 12, - 42; m = 0
27. P = 12, 42; m = -
3
4
31. P = 10, 32 ; slope undefined
28. P = 11, 32; m = -
2
5
32. P = 1 - 2, 02 ; slope undefined
In Problems 33–38, the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers may
vary. [Hint: It is not necessary to find the equation of the line. See Example 3.]
3
35. Slope - ; point 12, - 42
33. Slope 4; point 11, 22
34. Slope 2; point 1 - 2, 32
2
4
36. Slope ; point 1 - 3, 22
37. Slope - 2; point 1 - 2, - 32
38. Slope - 1; point 14, 12
3
In Problems 39–46, find an equation of the line L.
39.
40.
y
2
(2, 1)
L
(–2, 1) 2
–1
(–1, 3)
2
x
–2
–1
42.
y
3
(–1, 1)
x
2
L
y
–2
–2
2
L
(2, 2)
(1, 1)
(0, 0)
(0, 0)
–2
41. L
y
–1
2
x
–1
2
x
SECTION 2.3 Lines
y
43.
y
44.
3
y
3
45.
3
(1, 2)
(3, 3)
46.
–1
y = 2x
L
–1
3 x
L
–1
y = 2x
3 x
y = –x
(–1, 1)
3 x
–3
L
L is perpendicular
to y = 2x
L is parallel to y = –x
L is parallel to y = 2x
y
3
(1, 2)
L
179
1 x
y = –x
L is perpendicular
to y = –x
In Problems 47–72, find an equation for the line with the given properties. Express your answer using either the general form or the
slope–intercept form of the equation of a line, whichever you prefer.
47. Slope = 3; containing the point 1 - 2, 32
48. Slope = 2; containing the point 14, - 32
2
49. Slope = - ; containing the point 11, - 12
3
50. Slope =
1
; containing the point 13, 12
2
51. Containing the points 11, 32 and 1 - 1, 22
52. Containing the points 1 - 3, 42 and 12, 52
53. Slope = - 3; y@intercept = 3
54. Slope = - 2; y@intercept = - 2
55. x@intercept = 2; y@intercept = - 1
56. x@intercept = - 4; y@intercept = 4
57. Slope undefined; containing the point 12, 42
58. Slope undefined; containing the point 13, 82
59. Horizontal; containing the point 1 - 3, 22
60. Vertical; containing the point 14, - 52
61. Parallel to the line y = 2x; containing the point 1 - 1, 22
62. Parallel to the line y = - 3x; containing the point 1 - 1, 22
63. Parallel to the line 2x - y = - 2; containing the point 10, 02
64. Parallel to the line x - 2y = - 5; containing the point 10, 02
65. Parallel to the line x = 5; containing the point 14, 22
66. Parallel to the line y = 5; containing the point 14, 22
1
67. Perpendicular to the line y = x + 4; containing the point
2
11, - 22
68. Perpendicular to the line y = 2x - 3; containing the point
11, - 22
69. Perpendicular to the line 2x + y = 2; containing the point
1 - 3, 02
70. Perpendicular to the line x - 2y = - 5; containing the point
10, 42
71. Perpendicular to the line x = 8; containing the point 13, 42
72. Perpendicular to the line y = 8; containing the point 13, 42
In Problems 73–92, find the slope and y-intercept of each line. Graph the line.
1
73. y = 2x + 3
74. y = - 3x + 4
75. y = x - 1
2
1
79. x + 2y = 4
80. - x + 3y = 6
78. y = 2x +
2
85. x = - 4
84. x - y = 2
83. x + y = 1
88. x = 2
89. y - x = 0
90. x + y = 0
76.
1
x + y = 2
3
77. y =
1
x + 2
2
81. 2x - 3y = 6
82. 3x + 2y = 6
86. y = - 1
87. y = 5
91. 2y - 3x = 0
92. 3x + 2y = 0
In Problems 93–102, (a) find the intercepts of the graph of each equation and (b) graph the equation.
93. 2x + 3y = 6
94. 3x - 2y = 6
95. - 4x + 5y = 40
96. 6x - 4y = 24
97. 7x + 2y = 21
98. 5x + 3y = 18
99.
1
1
x + y = 1
2
3
100. x -
103. Find an equation of the x-axis.
2
y = 4
3
101. 0.2x - 0.5y = 1
102. - 0.3x + 0.4y = 1.2
104. Find an equation of the y-axis.
In Problems 105–108, the equations of two lines are given. Determine whether the lines are parallel, perpendicular, or neither.
1
108. y = - 2x + 3
105. y = 2x - 3
107. y = 4x + 5
106. y = x - 3
2
y = 2x + 4
y = - 4x + 2
1
y = - 2x + 4
y = - x + 2
2
180
CHAPTER 2 Graphs
In Problems 109–112, write an equation of each line. Express your answer using either the general form or the slope–intercept form of the
equation of a line, whichever you prefer.
109.
110.
4
26
111.
2
2
6
23
24
112.
3
22
23
2
3
22
23
3
22
Applications and Extensions
113. Geometry Use slopes to show that the triangle whose
vertices are 1 - 2, 52 , 11, 32 , and 1 - 1, 02 is a right triangle.
114. Geometry Use slopes to show that the quadrilateral
whose vertices are 11, - 12 , 14, 12 , 12, 22 , and 15, 42 is a
parallelogram.
115. Geometry Use slopes to show that the quadrilateral
whose vertices are 1 - 1, 02 , 12, 32 , 11, - 22 , and 14, 12 is a
rectangle.
116. Geometry Use slopes and the distance formula to show that
the quadrilateral whose vertices are 10, 02 , 11, 32 , 14, 22 ,
and 13, - 12 is a square.
117. Truck Rentals A truck rental company rents a moving
truck for one day by charging $39 plus $0.60 per mile. Write
a linear equation that relates the cost C, in dollars, of renting
the truck to the number x of miles driven. What is the cost
of renting the truck if the truck is driven 110 miles?
230 miles?
118. Cost Equation The fixed costs of operating a business are
the costs incurred regardless of the level of production.
Fixed costs include rent, fixed salaries, and costs of leasing
machinery. The variable costs of operating a business are
the costs that change with the level of output. Variable costs
include raw materials, hourly wages, and electricity. Suppose
that a manufacturer of jeans has fixed daily costs of $500
and variable costs of $8 for each pair of jeans manufactured.
Write a linear equation that relates the daily cost C, in
dollars, of manufacturing the jeans to the number x of jeans
manufactured. What is the cost of manufacturing 400 pairs
of jeans? 740 pairs?
119. Cost of Driving a Car The annual fixed costs of owning a
small sedan are $4462, assuming the car is completely paid
for. The cost to drive the car is approximately $0.17 per mile.
Write a linear equation that relates the cost C and the
number x of miles driven annually.
Source: AAA, April 2013
120. Wages of a Car Salesperson Dan receives $375 per week for
selling new and used cars at a car dealership in Oak Lawn,
Illinois. In addition, he receives 5% of the profit on any sales
that he generates. Write a linear equation that represents
Dan’s weekly salary S when he has sales that generate a
profit of x dollars.
121. Electricity Rates in Illinois Commonwealth Edison
Company supplies electricity to residential customers for
a monthly customer charge of $15.37 plus 8.21 cents per
kilowatt-hour for up to 800 kilowatt-hours (kW-hr).
(a) Write a linear equation that relates the monthly charge
C, in dollars, to the number x of kilowatt-hours used in a
month, 0 … x … 800.
(b) Graph this equation.
(c) What is the monthly charge for using 200 kilowatthours?
(d) What is the monthly charge for using 500 kilowatthours?
(e) Interpret the slope of the line.
Source: Commonwealth Edison Company, January 2014.
122. Electricity Rates in Florida Florida Power & Light Company
supplies electricity to residential customers for a monthly
customer charge of $7.24 plus 9.07 cents per kilowatt-hour
for up to 1000 kilowatt-hours.
(a) Write a linear equation that relates the monthly charge
C, in dollars, to the number x of kilowatt-hours used in a
month, 0 … x … 1000.
(b) Graph this equation.
(c) What is the monthly charge for using 200 kilowatthours?
(d) What is the monthly charge for using 500 kilowatthours?
(e) Interpret the slope of the line.
Source: Florida Power & Light Company, March 2014.
123. Measuring Temperature The relationship between Celsius
(°C) and Fahrenheit (°F) degrees of measuring temperature
is linear. Find a linear equation relating °C and °F if 0°C
corresponds to 32°F and 100°C corresponds to 212°F. Use
the equation to find the Celsius measure of 70°F.
124. Measuring Temperature The Kelvin (K) scale for measuring
temperature is obtained by adding 273 to the Celsius
temperature.
SECTION 2.3 Lines
(a) Write a linear equation relating K and °C.
(b) Write a linear equation relating K and °F (see
Problem 123).
125. Access Ramp A wooden access ramp is being built to reach
a platform that sits 30 inches above the floor. The ramp
drops 2 inches for every 25-inch run.
y
Platform
30 in.
Ramp
x
(a) Write a linear equation that relates the height y of the
ramp above the floor to the horizontal distance x from
the platform.
(b) Find and interpret the x-intercept of the graph of your
equation.
(c) Design requirements stipulate that the maximum run
be 30 feet and that the maximum slope be a drop of
1 inch for each 12 inches of run. Will this ramp meet the
requirements? Explain.
(d) What slopes could be used to obtain the 30-inch rise and
still meet design requirements?
Source: www.adaptiveaccess.com/wood_ramps.php
126. Cigarette Use A report in the Child Trends DataBase
indicated that in 2000, 20.6% of twelfth grade students
reported daily use of cigarettes. In 2012, 9.3% of twelfth
grade students reported daily use of cigarettes.
(a) Write a linear equation that relates the percent y of
twelfth grade students who smoke cigarettes daily to the
number x of years after 2000.
181
(b) Find the intercepts of the graph of your equation.
(c) Do these intercepts have meaningful interpretation?
(d) Use your equation to predict the percent for the year
2025. Is this result reasonable?
Source: www.childtrendsdatabank.org
127. Product Promotion A cereal company finds that the
number of people who will buy one of its products in the
first month that the product is introduced is linearly related
to the amount of money it spends on advertising. If it spends
$40,000 on advertising, then 100,000 boxes of cereal will be
sold, and if it spends $60,000, then 200,000 boxes will be
sold.
(a) Write a linear equation that relates the amount A spent
on advertising to the number x of boxes the company
aims to sell.
(b) How much expenditure on advertising is needed to sell
300,000 boxes of cereal?
(c) Interpret the slope.
128. Show that the line containing the points 1a, b2 and 1b, a2,
a ≠ b, is perpendicular to the line y = x. Also show that
the midpoint of 1a, b2 and 1b, a2 lies on the line y = x.
129. The equation 2x - y = C defines a family of lines, one line
for each value of C. On one set of coordinate axes, graph the
members of the family when C = - 4, C = 0, and C = 2.
Can you draw a conclusion from the graph about each
member of the family?
130. Prove that if two nonvertical lines have slopes whose
product is - 1, then the lines are perpendicular. [Hint:
Refer to Figure 47 and use the converse of the Pythagorean
Theorem.]
Explaining Concepts: Discussion and Writing
131. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 2x + 3y = 6
y
(b) - 2x + 3y = 6
(c) 3x - 4y = - 12
(d) x - y = 1
(e) x - y = - 1
x
(f) y = 3x - 5
(g) y = 2x + 3
(h) y = - 3x + 3
132. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 2x + 3y = 6
y
(b) 2x - 3y = 6
(c) 3x + 4y = 12
(d) x - y = 1
x
(e) x - y = - 1
(f) y = - 2x - 1
1
(g) y = - x + 10
2
(h) y = x + 4
133. The figure shows the graph of two parallel lines. Which of the
following pairs of equations might have such a graph?
(a) x - 2y = 3
y
x + 2y = 7
(b) x + y = 2
x + y = -1
x
(c) x - y = - 2
x - y = 1
(d) x - y = - 2
2x - 2y = - 4
(e) x + 2y = 2
x + 2y = - 1
134. The figure shows the graph of two perpendicular lines.
Which of the following pairs of equations might have such a
graph?
(a) y - 2x = 2
y
y + 2x = - 1
(b) y - 2x = 0
2y + x = 0
x
(c) 2y - x = 2
2y + x = - 2
(d) y - 2x = 2
x + 2y = - 1
(e) 2x + y = - 2
2y + x = - 2
135. m is for Slope The accepted symbol used to denote the
slope of a line is the letter m. Investigate the origin of this
practice. Begin by consulting a French dictionary and looking
up the French word monter. Write a brief essay on your
findings.
182
CHAPTER 2 Graphs
136. Grade of a Road The term grade is used to describe the
inclination of a road. How is this term related to the notion
of slope of a line? Is a 4% grade very steep? Investigate
the grades of some mountainous roads and determine their
slopes. Write a brief essay on your findings.
138. Can the equation of every line be written in slope–intercept
form? Why?
139. Does every line have exactly one x-intercept and one
y-intercept? Are there any lines that have no intercepts?
140. What can you say about two lines that have equal slopes and
equal y-intercepts?
141. What can you say about two lines with the same x-intercept
and the same y-intercept? Assume that the x-intercept is
not 0.
Steep
7% Grade
142. If two distinct lines have the same slope but different
x-intercepts, can they have the same y-intercept?
143. If two distinct lines have the same y-intercept but different
slopes, can they have the same x-intercept?
144. Which form of the equation of a line do you prefer to use?
Justify your position with an example that shows that your
choice is better than another. Have reasons.
137. Carpentry Carpenters use the term pitch to describe the
steepness of staircases and roofs. How is pitch related to
slope? Investigate typical pitches used for stairs and for
roofs. Write a brief essay on your findings.
145. What Went Wrong? A student is asked to find the slope of
the line joining ( - 3, 2) and (1, - 4). He states that the slope
3
is . Is he correct? If not, what went wrong?
2
Retain Your Knowledge
Problems 146–149 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
146. Simplify a
x2y -3
b . Assume x ≠ 0 and y ≠ 0. Express the
-2
x4y5
answer so that all exponents are positive.
147. The lengths of the legs of a right triangle are a = 8 and
b = 15. Find the hypotenuse.
148. Solve the equation: (x - 3)2 + 25 = 49
149. Solve 2x - 5 + 7 6 10. Express the answer using set
notation or interval notation. Graph the solution set.
2.4 Circles
PREPARING FOR THIS SECTION Before getting started, review the following:
r Completing the Square (Chapter R, Section R.5, p. 56)
r Square Root Method (Section 1.2, pp. 94–95)
Now Work the ‘Are You Prepared?’ problems on page 186.
OBJECTIVES 1 Write the Standard Form of the Equation of a Circle (p. 182)
2 Graph a Circle (p. 183)
3 Work with the General Form of the Equation of a Circle (p. 184)
1 Write the Standard Form of the Equation of a Circle
One advantage of a coordinate system is that it enables us to translate a geometric
statement into an algebraic statement, and vice versa. Consider, for example, the
following geometric statement that defines a circle.
DEFINITION
A circle is a set of points in the xy-plane that are a fixed distance r from a fixed
point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2
is called the center of the circle.
SECTION 2.4 Circles
y
183
Figure 49 shows the graph of a circle. To find the equation, let 1x, y2 represent
the coordinates of any point on a circle with radius r and center 1h, k2 . Then the
distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the
distance formula,
(x, y)
r
2 1x - h2 2 + 1y - k2 2 = r
(h, k )
x
or, equivalently,
1x - h2 2 + 1y - k2 2 = r 2
Figure 49 1x - h2 2 + 1y - k2 2 = r 2
DEFINITION
The standard form of an equation of a circle with radius r and center 1h, k2 is
1x - h2 2 + 1y - k2 2 = r 2
THEOREM
(1)
The standard form of an equation of a circle of radius r with center at the
origin 10, 02 is
x2 + y 2 = r 2
DEFINITION
If the radius r = 1, the circle whose center is at the origin is called the unit
circle and has the equation
x2 + y 2 = 1
See Figure 50. Notice that the graph of the unit circle is symmetric with respect to
the x-axis, the y-axis, and the origin.
y
1
1
Figure 50
Unit circle x2 + y2 = 1
EXAMPL E 1
Solution
(0,0)
1
x
1
Writing the Standard Form of the Equation of a Circle
Write the standard form of the equation of the circle with radius 5 and center 1 - 3, 62.
Substitute the values r = 5, h = - 3, and k = 6 into equation (1).
1x - h2 2 + 1y - k2 2 = r 2
1x + 32 2 + 1y - 62 2 = 25
Now Work
PROBLEM
9
r
2 Graph a Circle
EXAMPL E 2
Solution
Graphing a Circle
Graph the equation: 1x + 32 2 + 1y - 22 2 = 16
Since the equation is in the form of equation (1), its graph is a circle. To graph the
equation, compare the given equation to the standard form of the equation of a
circle. The comparison yields information about the circle.
184
CHAPTER 2 Graphs
(–3, 6)
1x + 32 2 + 1y - 22 2 = 16
y
1x - 1 - 32 2 2 + 1y - 22 2 = 42
6
4
(–7, 2)
–10
(–3, 2)
c
(1, 2)
2 x
–5
(–3, –2)
Figure 51 1x + 32 2 + 1y - 22 2 = 16
EX AMPLE 3
Solution
c
c
1x - h2 2 + 1y - k2 2 = r 2
We see that h = - 3, k = 2, and r = 4. The circle has center 1 - 3, 22 and a
radius of 4 units. To graph this circle, first plot the center 1 - 3, 22 . Since the radius
is 4, we can locate four points on the circle by plotting points 4 units to the left, to
the right, up, and down from the center. These four points can then be used as guides
to obtain the graph. See Figure 51.
r
Now Work
PROBLEMS
25(a)
AND
(b)
Finding the Intercepts of a Circle
For the circle 1x + 32 2 + 1y - 22 2 = 16, find the intercepts, if any, of its graph.
This is the equation discussed and graphed in Example 2. To find the x-intercepts, if
any, let y = 0. Then
1x + 32 2 + 1y - 22 2 = 16
In Words
The symbol { is read “plus or
minus.” It means to add and
subtract the quantity following
the { symbol. For example,
5 { 2 means “5 - 2 = 3 or
5 + 2 = 7.”
1x + 32 2 + 10 - 22 2 = 16
y = 0
1x + 32 + 4 = 16
2
Simplify.
1x + 32 = 12
2
Simplify.
x + 3 = { 212
Apply the Square Root Method.
x = - 3 { 223
Solve for x.
The x-intercepts are - 3 - 223 ≈ - 6.46 and - 3 + 223 ≈ 0.46.
To find the y-intercepts, if any, let x = 0. Then
1x + 32 2 + 1y - 22 2 = 16
10 + 32 2 + 1y - 22 2 = 16
9 + 1y - 22 2 = 16
1y - 22 2 = 7
y - 2 = { 27
y = 2 { 27
The y-intercepts are 2 - 27 ≈ - 0.65 and 2 + 27 ≈ 4.65.
Look back at Figure 51 to verify the approximate locations of the intercepts.
Now Work
PROBLEM
25(C)
r
3 Work with the General Form of the Equation of a Circle
If we eliminate the parentheses from the standard form of the equation of the circle
given in Example 2, we get
1x + 32 2 + 1y - 22 2 = 16
x2 + 6x + 9 + y2 - 4y + 4 = 16
which simplifies to
x2 + y2 + 6x - 4y - 3 = 0
(2)
It can be shown that any equation of the form
x2 + y2 + ax + by + c = 0
has a graph that is a circle, is a point, or has no graph at all. For example, the graph of
the equation x2 + y2 = 0 is the single point 10, 02. The equation x2 + y2 + 5 = 0, or
x2 + y2 = - 5, has no graph, because sums of squares of real numbers are never negative.
SECTION 2.4 Circles
DEFINITION
185
When its graph is a circle, the equation
x2 + y2 + ax + by + c = 0
is the general form of the equation of a circle.
Now Work
PROBLEM
15
If an equation of a circle is in general form, we use the method of completing the
square to put the equation in standard form so that we can identify its center and radius.
EXAMPL E 4
Graphing a Circle Whose Equation Is in General Form
Graph the equation: x2 + y2 + 4x - 6y + 12 = 0
Solution
Group the terms involving x, group the terms involving y, and put the constant on
the right side of the equation. The result is
1x2 + 4x2 + 1y2 - 6y2 = - 12
y
(–2, 4)
4
1
(–3, 3)
Next, complete the square of each expression in parentheses. Remember that any
number added on the left side of the equation must also be added on the right.
1x2 + 4x + 42 + 1y2 - 6y + 92 = - 12 + 4 + 9
(–1, 3)
c
6
4 2
a b = 4
2
(–2, 3)
(–2, 2)
c
6
a
-6 2
b = 9
2
1x + 22 2 + 1y - 32 2 = 1 Factor.
1 x
–3
Figure 52 1x + 22 2 + 1y - 32 2 = 1
This equation is the standard form of the equation of a circle with radius 1 and
center 1 - 2, 32 . To graph the equation, use the center 1 - 2, 32 and the radius 1. See
Figure 52.
Now Work
EXAMPL E 5
PROBLEM
29
r
Using a Graphing Utility to Graph a Circle
Graph the equation: x2 + y2 = 4
Solution
Y1 5 √42x 2
This is the equation of a circle with center at the origin and radius 2. To graph this
equation, solve for y.
x2 + y 2 = 4
y 2 = 4 - x2
2.5
y = { 24 - x
24
4
22.5
Y2 5 2√42x 2
Figure 53 x + y = 4
2
2
Subtract x 2 from each side.
2
Apply the Square Root Method to solve for y.
There are two equations to graph: first graph Y1 = 24 - x2 and then graph
Y2 = - 24 - x2 on the same square screen. (Your circle will appear oval if you do
not use a square screen.*) See Figure 53.
Overview
The discussion in Sections 2.3 and 2.4 about lines and circles dealt with two main
types of problems that can be generalized as follows:
1. Given an equation, classify it and graph it.
2. Given a graph, or information about a graph, find its equation.
This text deals with both types of problems. We shall study various equations,
classify them, and graph them. The second type of problem is usually more difficult
to solve than the first.
*The square screen ratio for the TI-84 Plus C calculator is 8:5.
186
CHAPTER 2 Graphs
2.4 Assess Your Understanding
Are You Prepared? Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. To complete the square of x2 + 10x, you would
(add/subtract) the number
. (p. 56)
2. Use the Square Root Method to solve the equation
1x - 22 2 = 9. (pp. 94–95)
Concepts and Vocabulary
3. True or False Every equation of the form
7. Choose the equation of a circle with radius 6 and center (3, –5).
(a) 1x - 32 2 + 1y + 52 2 = 6
(b) 1x + 32 2 + 1y - 52 2 = 36
(c) 1x + 32 2 + 1y - 52 2 = 6
(d) 1x - 32 2 + 1y + 52 2 = 36
x + y + ax + by + c = 0
2
2
has a circle as its graph.
4. For a circle, the
any point on the circle.
is the distance from the center to
8. The equation of a circle can be changed from general form
to standard from by doing which of the following?
(a) completing the squares
(b) solving for x
(c) solving for y
(d) squaring both sides
5. True or False The radius of the circle x2 + y2 = 9 is 3.
6. True or False The center of the circle
1x + 32 2 + 1y - 22 2 = 13
is (3, - 2).
Skill Building
In Problems 9–12, find the center and radius of each circle. Write the standard form of the equation.
10.
9. y
11. y
y
(4, 2)
12. y
(2, 3)
(1, 2)
(0, 1)
(2, 1)
(0, 1)
(1, 2)
x
(1, 0)
x
x
x
In Problems 13–22, write the standard form of the equation and the general form of the equation of each circle of radius r and center 1h, k2.
Graph each circle.
13. r = 2;
17. r = 5;
21. r =
1
;
2
1h, k2 = 10, 02
1h, k2 = 14, - 32
14. r = 3;
18. r = 4;
1h, k2 = 10, 02
1h, k2 = 12, - 32
1
1h, k2 = a , 0b
2
15. r = 2;
19. r = 4;
22. r =
1
;
2
1h, k2 = 10, 22
1h, k2 = 1 - 2, 12
16. r = 3;
20. r = 7;
1h, k2 = 11, 02
1h, k2 = 1 - 5, - 22
1
1h, k2 = a0, - b
2
In Problems 23–36, (a) find the center 1h, k2 and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any.
23. x2 + y2 = 4
24. x2 + 1y - 12 2 = 1
25. 21x - 32 2 + 2y2 = 8
26. 31x + 12 2 + 31y - 12 2 = 6
27. x2 + y2 - 2x - 4y - 4 = 0
28. x2 + y2 + 4x + 2y - 20 = 0
29. x2 + y2 + 4x - 4y - 1 = 0
1
32. x2 + y2 + x + y - = 0
2
35. 2x2 + 8x + 2y2 = 0
30. x2 + y2 - 6x + 2y + 9 = 0
31. x2 + y2 - x + 2y + 1 = 0
33. 2x2 + 2y2 - 12x + 8y - 24 = 0
34. 2x2 + 2y2 + 8x + 7 = 0
36. 3x2 + 3y2 - 12y = 0
In Problems 37–44, find the standard form of the equation of each circle.
37. Center at the origin and containing the point 1 - 2, 32
38. Center 11, 02 and containing the point 1 - 3, 22
39. Center 12, 32 and tangent to the x-axis
40. Center 1 - 3, 12 and tangent to the y-axis
43. Center 1 - 1, 32 and tangent to the line y = 2
44. Center 14, - 22 and tangent to the line x = 1
41. With endpoints of a diameter at 11, 42 and 1 - 3, 22
42. With endpoints of a diameter at 14, 32 and 10, 12
SECTION 2.4 Circles
187
In Problems 45–48, match each graph with the correct equation.
(a) 1x - 32 2 + 1y + 32 2 = 9 (b) 1x + 12 2 + 1y - 22 2 = 4 (c) 1x - 12 2 + 1y + 22 2 = 4 (d) 1x + 32 2 + 1y - 32 2 = 9
45.
46.
47.
4
6
26.4
6.4 29.6
48.
4
6
9.6 26.4
24
26
6.4 29.6
9.6
24
26
Applications and Extensions
53. Weather Satellites Earth is represented on a map of a
portion of the solar system so that its surface is the circle
with equation x2 + y2 + 2x + 4y - 4091 = 0. A weather
satellite circles 0.6 unit above Earth with the center of its
circular orbit at the center of Earth. Find the equation for
the orbit of the satellite on this map.
49. Find the area of the square in the figure.
y
x2 y2 9
x
r
50. Find the area of the blue shaded region in the figure, assuming
the quadrilateral inside the circle is a square.
y
x 2 y 2 36
54. The tangent line to a circle may be defined as the line that
intersects the circle in a single point, called the point of
tangency. See the figure.
x
y
r
51. Ferris Wheel The original Ferris wheel was built in 1893 by
Pittsburgh, Pennsylvania bridge builder George W. Ferris.
The Ferris wheel was originally built for the 1893 World’s
Fair in Chicago, but it was also later reconstructed for the
1904 World’s Fair in St. Louis. It had a maximum height of
264 feet and a wheel diameter of 250 feet. Find an equation
for the wheel if the center of the wheel is on the y-axis.
Source: guinnessworldrecords.com
52. Ferris Wheel Opening in 2014 in Las Vegas, The High
Roller observation wheel has a maximum height of 550 feet
and a diameter of 520 feet, with one full rotation taking
approximately 30 minutes. Find an equation for the wheel if
the center of the wheel is on the y-axis.
Source: Las Vegas Review Journal
x
If the equation of the circle is x2 + y2 = r 2 and the equation
of the tangent line is y = mx + b, show that:
(a) r 2 11 + m2 2 = b2
[Hint: The quadratic equation x2 + 1mx + b2 2 = r 2
has exactly one solution.]
- r 2m r 2
(b) The point of tangency is ¢
, ≤.
b
b
(c) The tangent line is perpendicular to the line containing
the center of the circle and the point of tangency.
55. The Greek Method The Greek method for finding the
equation of the tangent line to a circle uses the fact that at
any point on a circle, the lines containing the center and the
tangent line are perpendicular (see Problem 54). Use this
method to find an equation of the tangent line to the circle
x2 + y2 = 9 at the point 11, 2122 .
56. Use the Greek method described in Problem 55
to find an equation of the tangent line to the circle
x2 + y2 - 4x + 6y + 4 = 0 at the point 13, 212 - 32 .
188
CHAPTER 2 Graphs
57. Refer to Problem 54. The line x - 2y + 4 = 0 is tangent to
a circle at 10, 22 . The line y = 2x - 7 is tangent to the same
circle at 13, - 12 . Find the center of the circle.
59. If a circle of radius 2 is made to roll along the x-axis, what is
an equation for the path of the center of the circle?
60. If the circumference of a circle is 6p, what is its radius?
58. Find an equation of the line containing the centers of the
two circles
x2 + y2 - 4x + 6y + 4 = 0
and
x2 + y2 + 6x + 4y + 9 = 0
Explaining Concepts: Discussion and Writing
61. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 1x - 22 2 + 1y + 32 2 = 13
y
(b) 1x - 22 2 + 1y - 22 2 = 8
(c) 1x - 22 2 + 1y - 32 2 = 13
(d) 1x + 22 2 + 1y - 22 2 = 8
(e) x2 + y2 - 4x - 9y = 0
x
(f) x2 + y2 + 4x - 2y = 0
(g) x2 + y2 - 9x - 4y = 0
(h) x2 + y2 - 4x - 4y = 4
62. Which of the following equations might have the graph
shown? (More than one answer is possible.)
(a) 1x - 22 2 + y2 = 3
y
(b) 1x + 22 2 + y2 = 3
(c) x2 + 1y - 22 2 = 3
(d) 1x + 22 2 + y2 = 4
(e) x2 + y2 + 10x + 16 = 0
x
(f) x2 + y2 + 10x - 2y = 1
(g) x2 + y2 + 9x + 10 = 0
(h) x2 + y2 - 9x - 10 = 0
63. Explain how the center and radius of a circle can be used to graph the circle.
64. What Went Wrong? A student stated that the center and radius of the graph whose equation is (x + 3)2 + (y - 2)2 = 16 are
(3, - 2) and 4, respectively. Why is this incorrect?
Retain Your Knowledge
Problems 65–68 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
65. Find the area and circumference of a circle of radius 13 cm.
2
66. Multiply (3x - 2)(x - 2x + 3). Express the answer as a
polynomial in standard form.
68. Aaron can load a delivery van in 22 minutes. Elizabeth can
load the same van in 28 minutes. How long would it take them
to load the van if they worked together?
67. Solve the equation: 22x2 + 3x - 1 = x + 1
‘Are You Prepared?’ Answers
1. add; 25
2. 5 - 1, 56
2.5 Variation
OBJECTIVES 1 Construct a Model Using Direct Variation (p. 189)
2 Construct a Model Using Inverse Variation (p. 189)
3 Construct a Model Using Joint Variation or Combined Variation (p. 190)
When a mathematical model is developed for a real-world problem, it often involves
relationships between quantities that are expressed in terms of proportionality:
Force is proportional to acceleration.
When an ideal gas is held at a constant temperature, pressure and volume are
inversely proportional.
The force of attraction between two heavenly bodies is inversely proportional
to the square of the distance between them.
Revenue is directly proportional to sales.
SECTION 2.5 Variation
189
Each of the preceding statements illustrates the idea of variation, or how one quantity
varies in relation to another quantity. Quantities may vary directly, inversely, or jointly.
1 Construct a Model Using Direct Variation
DEFINITION
Let x and y denote two quantities. Then y varies directly with x, or y is directly
proportional to x, if there is a nonzero number k such that
y = kx
The number k is called the constant of proportionality.
y
The graph in Figure 54 illustrates the relationship between y and x if y varies
directly with x and k 7 0, x Ú 0. Note that the constant of proportionality is, in fact,
the slope of the line.
If two quantities vary directly, then knowing the value of each quantity in one
instance enables us to write a formula that is true in all cases.
x
Figure 54 y = k x ; k 7 0, x Ú 0
Mortgage Payments
EXAMPL E 1
The monthly payment p on a mortgage varies directly with the amount borrowed B.
If the monthly payment on a 30-year mortgage is $6.65 for every $1000 borrowed,
find a formula that relates the monthly payment p to the amount borrowed B for
a mortgage with these terms. Then find the monthly payment p when the amount
borrowed B is $120,000.
Solution
Because p varies directly with B, we know that
p = kB
p
for some constant k. Because p = 6.65 when B = 1000, it follows that
Monthly Payment
800
6.65 = k 110002
(120 000, 798)
600
k = 0.00665
Since p = kB,
400
p = 0.00665B
200
0
Solve for k.
40
80
120
160
Amount Borrowed (000's)
B
In particular, when B = $120,000,
p = 0.006651$120,0002 = $798
Figure 55 illustrates the relationship between the monthly payment p and the
amount borrowed B.
Figure 55
Now Work
PROBLEMS
5
AND
23
r
2 Construct a Model Using Inverse Variation
y
DEFINITION
Let x and y denote two quantities. Then y varies inversely with x, or y is
inversely proportional to x, if there is a nonzero constant k such that
y =
k
x
x
Figure 56 y =
k
; k 7 0, x 7 0
x
The graph in Figure 56 illustrates the relationship between y and x if y varies
inversely with x and k 7 0, x 7 0.
190
CHAPTER 2 Graphs
EX AMPLE 2
Maximum Weight That Can Be Supported by a Piece of Pine
See Figure 57. The maximum weight W that can be safely supported by a 2-inch by
4-inch piece of pine varies inversely with its length l. Experiments indicate that the
maximum weight that a 10-foot-long 2-by-4 piece of pine can support is 500 pounds.
Write a general formula relating the maximum weight W (in pounds) to length l (in
feet). Find the maximum weight W that can be safely supported by a length of 25 feet.
Solution Because W varies inversely with l, we know that
W =
k
l
for some constant k. Because W = 500 when l = 10, we have
k
10
500 =
k = 5000
Since W =
k
,
l
W =
Figure 57
5000
l
In particular, the maximum weight W that can be safely supported by a piece of pine
25 feet in length is
W
600
5000
= 200 pounds
25
W =
(10, 500)
500
400
Figure 58 illustrates the relationship between the weight W and the length l.
300
Now Work
200
(25, 200)
PROBLEM
r
33
100
0
5
10
Figure 58 W =
15
20
25
l
5000
l
3 Construct a Model Using Joint Variation
or Combined Variation
When a variable quantity Q is proportional to two or more other variables, we say
that Q varies jointly with these quantities. Finally, combinations of direct and/or
inverse variation may occur. This is usually referred to as combined variation.
EX AMPLE 3
Loss of Heat through a Wall
The loss of heat through a wall varies jointly with the area of the wall and the
difference between the inside and outside temperatures and varies inversely with
the thickness of the wall. Write an equation that relates these quantities.
Solution
Begin by assigning symbols to represent the quantities:
L = Heat loss
T = Temperature difference
A = Area of wall
d = Thickness of wall
Then
L = k
where k is the constant of proportionality.
AT
d
r
SECTION 2.5 Variation
191
In direct or inverse variation, the quantities that vary may be raised to powers.
For example, in the early seventeenth century, Johannes Kepler (1571–1630) discovered
that the square of the period of revolution T of a planet around the Sun varies
directly with the cube of its mean distance a from the Sun. That is, T 2 = ka3, where
k is the constant of proportionality.
EXAMPL E 4
Force of the Wind on a Window
The force F of the wind on a flat surface positioned at a right angle to the direction
of the wind varies jointly with the area A of the surface and the square of the speed v
of the wind. A wind of 30 miles per hour blowing on a window measuring 4 feet by 5
feet has a force of 150 pounds. See Figure 59. What force does a wind of 50 miles per
hour exert on a window measuring 3 feet by 4 feet?
Solution
Since F varies jointly with A and v2, we have
F = kAv2
where k is the constant of proportionality. We are told that F = 150 when
A = 4 # 5 = 20 and v = 30. Then
150 = k 1202 19002 F = kAv 2, F = 150, A = 20, v = 30
Wind
1
120
k =
Figure 59
Since F = kAv2,
F =
1
Av2
120
For a wind of 50 miles per hour blowing on a window whose area is A = 3 # 4 = 12
square feet, the force F is
F =
Now Work
1
1122 125002 = 250 pounds
120
PROBLEM
r
41
2.5 Assess Your Understanding
Concepts and Vocabulary
1. If x and y are two quantities, then y is directly proportional
to x if there is a nonzero number k such that
.
k
2. True or False If y varies directly with x, then y = , where
x
k is a constant.
3. Which equation represents a joint variation model?
(a) y = 5x
(b) y = 5xzw
5
5xz
(c) y =
(d) y =
x
w
kx
4. Choose the best description for the model y =
, if k is a
z
nonzero constant.
(a) y varies jointly with x and z.
(b) y is inversely proportional to x and z.
(c) y varies directly with x and inversely with z.
(d) y is directly proportion to z and inversely proportional
to x.
Skill Building
In Problems 5–16, write a general formula to describe each variation.
5. y varies directly with x; y = 2 when x = 10
6. v varies directly with t; v = 16 when t = 2
7. A varies directly with x2; A = 4p when x = 2
8. V varies directly with x3; V = 36p when x = 3
9. F varies inversely with d 2; F = 10 when d = 5
10. y varies inversely with 1x; y = 4 when x = 9
11. z varies directly with the sum of the squares of x and
y; z = 5 when x = 3 and y = 4
12. T varies jointly with the cube root of x and the square of d;
T = 18 when x = 8 and d = 3
192
CHAPTER 2 Graphs
13. M varies directly with the square of d and inversely with the square root of x; M = 24 when x = 9 and d = 4
14. z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y = 3
15. The square of T varies directly with the cube of a and inversely with the square of d; T = 2 when a = 2 and d = 4
16. The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y = 4
Applications and Extensions
In Problems 17–22, write an equation that relates the quantities.
17. Geometry The volume V of a sphere varies directly with the
4p
cube of its radius r. The constant of proportionality is
.
3
18. Geometry The square of the length of the hypotenuse c of a
right triangle varies jointly with the sum of the squares of the
lengths of its legs a and b. The constant of proportionality
is 1.
19. Geometry The area A of a triangle varies jointly with the
product of the lengths of the base b and the height h. The
1
constant of proportionality is .
2
20. Geometry The perimeter p of a rectangle varies jointly with
the sum of the lengths of its sides l and w. The constant of
proportionality is 2.
21. Physics: Newton’s Law The force F (in newtons) of
attraction between two bodies varies jointly with their masses
m and M (in kilograms) and inversely with the square of
the distance d (in meters) between them. The constant of
proportionality is G = 6.67 * 10-11.
22. Physics: Simple Pendulum The period of a pendulum is the
time required for one oscillation; the pendulum is usually
referred to as simple when the angle made to the vertical is
less than 5°. The period T of a simple pendulum (in seconds)
varies directly with the square root of its length l (in feet).
2p
The constant of proportionality is
.
232
23. Mortgage Payments The monthly payment p on a mortgage
varies directly with the amount borrowed B. If the monthly
payment on a 30-year mortgage is $6.49 for every $1000
borrowed, find a linear equation that relates the monthly
payment p to the amount borrowed B for a mortgage with
these terms. Then find the monthly payment p when the
amount borrowed B is $145,000.
24. Mortgage Payments The monthly payment p on a mortgage
varies directly with the amount borrowed B. If the monthly
payment on a 15-year mortgage is $8.99 for every $1000
borrowed, find a linear equation that relates the monthly
payment p to the amount borrowed B for a mortgage with
these terms. Then find the monthly payment p when the
amount borrowed B is $175,000.
25. Physics: Falling Objects The distance s that an object falls is
directly proportional to the square of the time t of the fall.
If an object falls 16 feet in 1 second, how far will it fall in
3 seconds? How long will it take an object to fall 64 feet?
26. Physics: Falling Objects The velocity v of a falling object
is directly proportional to the time t of the fall. If, after
2 seconds, the velocity of the object is 64 feet per second,
what will its velocity be after 3 seconds?
27. Physics: Stretching a Spring The elongation E of a spring
balance varies directly with the applied weight W (see the
figure). If E = 3 when W = 20, find E when W = 15.
E
W
28. Physics: Vibrating String The rate of vibration of a string
under constant tension varies inversely with the length of
the string. If a string is 48 inches long and vibrates 256 times
per second, what is the length of a string that vibrates 576
times per second?
29. Revenue Equation At the corner Shell station, the revenue
R varies directly with the number g of gallons of gasoline
sold. If the revenue is $47.40 when the number of gallons
sold is 12, find a linear equation that relates revenue R to the
number g of gallons of gasoline sold. Then find the revenue
R when the number of gallons of gasoline sold is 10.5.
30. Cost Equation The cost C of roasted almonds varies
directly with the number A of pounds of almonds purchased.
If the cost is $23.75 when the number of pounds of roasted
almonds purchased is 5, find a linear equation that relates
the cost C to the number A of pounds of almonds purchased.
Then find the cost C when the number of pounds of almonds
purchased is 3.5.
31. Demand Suppose that the demand D for candy at the
movie theater is inversely related to the price p.
(a) When the price of candy is $2.75 per bag, the theater
sells 156 bags of candy. Express the demand for candy in
terms of its price.
(b) Determine the number of bags of candy that will be sold
if the price is raised to $3 a bag.
32. Driving to School The time t that it takes to get to school
varies inversely with your average speed s.
(a) Suppose that it takes you 40 minutes to get to school
when your average speed is 30 miles per hour.
Express the driving time to school in terms of average
speed.
(b) Suppose that your average speed to school is 40 miles
per hour. How long will it take you to get to school?
33. Pressure The volume of a gas V held at a constant
temperature in a closed container varies inversely with its
pressure P. If the volume of a gas is 600 cubic centimeters 1cm3 2
when the pressure is 150 millimeters of mercury (mm Hg), find
the volume when the pressure is 200 mm Hg.
SECTION 2.5 Variation
193
34. Resistance The current i in a circuit is inversely proportional
to its resistance Z measured in ohms. Suppose that when the
current in a circuit is 30 amperes, the resistance is 8 ohms.
Find the current in the same circuit when the resistance is
10 ohms.
surface and the square of the velocity of the wind. If the
force on an area of 20 square feet is 11 pounds when the
wind velocity is 22 miles per hour, find the force on a surface
area of 47.125 square feet when the wind velocity is 36.5
miles per hour.
35. Weight The weight of an object above the surface of Earth
varies inversely with the square of the distance from the
center of Earth. If Maria weighs 125 pounds when she is on
the surface of Earth (3960 miles from the center), determine
Maria’s weight when she is at the top of Mount McKinley
(3.8 miles from the surface of Earth).
41. Horsepower The horsepower (hp) that a shaft can safely
transmit varies jointly with its speed (in revolutions per
minute, rpm) and the cube of its diameter. If a shaft of a certain
material 2 inches in diameter can transmit 36 hp at 75 rpm,
what diameter must the shaft have in order to transmit 45 hp
at 125 rpm?
36. Weight of a Body The weight of a body above the surface of
Earth varies inversely with the square of the distance from
the center of Earth. If a certain body weighs 55 pounds when
it is 3960 miles from the center of Earth, how much will it
weigh when it is 3965 miles from the center?
42. Chemistry: Gas Laws The volume V of an ideal gas
varies directly with the temperature T and inversely with the
pressure P. Write an equation relating V, T, and P using k as
the constant of proportionality. If a cylinder contains oxygen at
a temperature of 300 K and a pressure of 15 atmospheres in a
volume of 100 liters, what is the constant of proportionality
k? If a piston is lowered into the cylinder, decreasing the
volume occupied by the gas to 80 liters and raising the
temperature to 310 K, what is the gas pressure?
37. Geometry The volume V of a right circular cylinder varies
jointly with the square of its radius r and its height h. The
constant of proportionality is p. See the figure. Write an
equation for V.
r
h
38. Geometry The volume V of a right circular cone varies
jointly with the square of its radius r and its height h. The
p
constant of proportionality is . See the figure. Write an
3
equation for V.
h
r
39. Intensity of Light The intensity I of light (measured in
foot-candles) varies inversely with the square of the distance
from the bulb. Suppose that the intensity of a 100-watt light
bulb at a distance of 2 meters is 0.075 foot-candle. Determine
the intensity of the bulb at a distance of 5 meters.
40. Force of the Wind on a Window The force exerted by the
wind on a plane surface varies jointly with the area of the
43. Physics: Kinetic Energy The kinetic energy K of a moving
object varies jointly with its mass m and the square of its
velocity v. If an object weighing 25 kilograms and moving
with a velocity of 10 meters per second has a kinetic energy
of 1250 joules, find its kinetic energy when the velocity is
15 meters per second.
44. Electrical Resistance of a Wire The electrical resistance of a
wire varies directly with the length of the wire and inversely with
the square of the diameter of the wire. If a wire 432 feet long
and 4 millimeters in diameter has a resistance of 1.24 ohms,
find the length of a wire of the same material whose resistance
is 1.44 ohms and whose diameter is 3 millimeters.
45. Measuring the Stress of Materials The stress in the material
of a pipe subject to internal pressure varies jointly with the
internal pressure and the internal diameter of the pipe and
inversely with the thickness of the pipe. The stress is 100
pounds per square inch when the diameter is 5 inches, the
thickness is 0.75 inch, and the internal pressure is 25 pounds
per square inch. Find the stress when the internal pressure is
40 pounds per square inch if the diameter is 8 inches and the
thickness is 0.50 inch.
46. Safe Load for a Beam The maximum safe load for a
horizontal rectangular beam varies jointly with the width of
the beam and the square of the thickness of the beam and
inversely with its length. If an 8-foot beam will support up
to 750 pounds when the beam is 4 inches wide and 2 inches
thick, what is the maximum safe load in a similar beam
10 feet long, 6 inches wide, and 2 inches thick?
Explaining Concepts: Discussion and Writing
47. In the early seventeenth century, Johannes Kepler discovered
that the square of the period T of the revolution of a planet
around the Sun varies directly with the cube of its mean
distance a from the Sun. Go to the library and research this
law and Kepler’s other two laws. Write a brief paper about
these laws and Kepler’s place in history.
48. Using a situation that has not been discussed in the text,
write a real-world problem that you think involves two
variables that vary directly. Exchange your problem with
another student’s to solve and critique.
49. Using a situation that has not been discussed in the text,
write a real-world problem that you think involves two
variables that vary inversely. Exchange your problem with
another student’s to solve and critique.
50. Using a situation that has not been discussed in the text,
write a real-world problem that you think involves three
variables that vary jointly. Exchange your problem with
another student’s to solve and critique.
194
CHAPTER 2 Graphs
Retain Your Knowledge
Problems 51–54 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
4 3>2
51. Factor 3x3 + 25x2 - 12x - 100 completely.
53. Simplify: a b
25
3
5
x - 2
.
54. Rationalize the denominator of
52. Add
and simplify the result.
+ 2
x + 3
x + 7x + 12
27 - 2
Chapter Review
Things to Know
Formulas
Distance formula (p. 151)
d = 21x2 - x1 2 2 + 1y2 - y1 2 2
Midpoint formula (p. 154)
1x, y2 = a
Slope (p. 167)
Parallel lines (p. 175)
x1 + x2 y1 + y2
,
b
2
2
y2 - y1
if x1 ≠ x2; undefined if x1 = x2
m =
x2 - x1
Equal slopes 1m1 = m2 2 and different y-intercepts 1b1 ≠ b2 2
Perpendicular lines (p. 176)
Product of slopes is - 1 1m1 # m2 = - 12
Direct variation (p. 189)
y = kx
Inverse variation (p. 189)
y =
k
x
Equations of Lines and Circles
Vertical line (p. 171)
x = a; a is the x-intercept
Horizontal line (p. 172)
y = b; b is the y-intercept
Point–slope form of the equation of a line (p. 171)
y - y1 = m1x - x1 2; m is the slope of the line, 1x1 , y1 2 is a point on the line
Slope–intercept form of the equation of a line (p. 172)
y = mx + b; m is the slope of the line, b is the y-intercept
General form of the equation of a line (p. 174)
Ax + By = C; A, B not both 0
Standard form of the equation of a circle (p. 183)
1x - h2 2 + 1y - k2 2 = r 2; r is the radius of the circle, 1h, k2 is the center
of the circle
Equation of the unit circle (p. 183)
x2 + y 2 = 1
General form of the equation of a circle (p. 185)
x2 + y2 + ax + by + c = 0, with restrictions on a, b, and c
Objectives
Section You should be able to …
Examples
Review Exercises
2.1
Use the distance formula (p. 151)
Use the midpoint formula (p. 153)
1–3
4
1(a)–3(a), 29, 30(a), 31
1(b)–3(b), 31
Graph equations by plotting points (p. 157)
Find intercepts from a graph (p. 159)
1–3
4
4
5
1
2
2.2
1
2
2.3
3
Find intercepts from an equation (p. 160)
5
6–10
4
6–9
6–10
5
Test an equation for symmetry with respect to the x-axis,
the y-axis, and the origin (p. 160)
Know how to graph key equations (p. 163)
10–12
26, 27
1
Calculate and interpret the slope of a line (p. 167)
1, 2
2
Graph lines given a point and the slope (p. 170)
3
1(c)–3(c), 1(d)–3(d),
32
28
3
Find the equation of a vertical line (p. 170)
4
17
Chapter Review
Section
You should be able to . . .
Examples
Review Exercises
Use the point–slope form of a line; identify horizontal lines (p. 171)
5, 6
16
5
Find the equation of a line given two points (p. 172)
7
18, 19
6
Write the equation of a line in slope–intercept form (p. 172)
8
16, 18–21
7
Identify the slope and y-intercept of a line from its equation (p. 173)
8
22, 23
8
Graph lines written in general form using intercepts (p. 174)
9
24, 25
9
Find equations of parallel lines (p. 175)
10, 11
20
Find equations of perpendicular lines (p. 176)
12, 13
21, 30(b)
2
Write the standard form of the equation of a circle (p. 182)
Graph a circle (p. 183)
1
2, 3, 5
11, 12, 31
13–15
3
Work with the general form of the equation of a circle (p. 184)
4
14, 15
1
2
Construct a model using direct variation (p. 189)
Construct a model using inverse variation (p. 189)
1
2
33
34
3
Construct a model using joint or combined variation (p. 190)
3, 4
35
4
10
2.4
2.5
1
195
Review Exercises
In Problems 1–3, find the following for each pair of points:
(a) The distance between the points
(b) The midpoint of the line segment connecting the points
(c) The slope of the line containing the points
(d) Interpret the slope found in part (c)
5. List the intercepts of the graph below.
y
1. 10, 02; 14, 22
2
2. 11, - 12; 1 - 2, 32
4
3. 14, - 42; 14, 82
4
x
2
4. Graph y = x2 + 4 by plotting points.
In Problems 6–10, list the intercepts and test for symmetry with respect to the x-axis, the y-axis, and the origin.
7. x2 + 4y2 = 16
6. 2x = 3y2
8. y = x4 + 2x2 + 1
10. x2 + x + y2 + 2y = 0
9. y = x3 - x
In Problems 11 and 12, find the standard form of the equation of the circle whose center and radius are given.
11. 1h, k2 = 1 - 2, 32; r = 4
12. 1h, k2 = 1 - 1, - 22; r = 1
In Problems 13–15, find the center and radius of each circle. Graph each circle. Find the intercepts, if any, of each circle.
13. x2 + 1y - 12 2 = 4
14. x2 + y2 - 2x + 4y - 4 = 0
15. 3x2 + 3y2 - 6x + 12y = 0
In Problems 16–21, find an equation of the line having the given characteristics. Express your answer using either the general form or the
slope–intercept form of the equation of a line, whichever you prefer.
16. Slope = - 2;
containing the point 13, - 12
18. y@intercept = - 2;
containing the point 15, - 32
20. Parallel to the line 2x - 3y = - 4;
21. Perpendicular to the line x + y = 2;
17. Vertical; containing the point 1 - 3, 42
19. Containing the points 13, - 42 and 12, 12
containing the point 1 - 5, 32
containing the point 14, - 32
In Problems 22 and 23, find the slope and y-intercept of each line. Graph the line, labeling any intercepts.
1
1
1
22. 4x - 5y = - 20
23. x - y = 2
3
6
In Problems 24 and 25, find the intercepts and graph each line.
1
1
24. 2x - 3y = 12
25. x + y = 2
2
3
196
CHAPTER 2 Graphs
26. Sketch a graph of y = x3.
27. Sketch a graph of y = 2x.
2
28. Graph the line with slope containing the point 11, 22 .
3
29. Show that the points A = 13, 42, B = 11, 12 , and
C = 1 - 2, 32 are the vertices of an isosceles triangle.
30. Show that the points A = 1 - 2, 02, B = 1 - 4, 42 , and
C = 18, 52 are the vertices of a right triangle in two ways:
(a) By using the converse of the Pythagorean Theorem
(b) By using the slopes of the lines joining the vertices
31. The endpoints of the diameter of a circle are 1 - 3, 22 and
15, - 62 . Find the center and radius of the circle. Write the
standard equation of this circle.
32. Show that the points A = 12, 52, B = 16, 12 ,
C = 18, - 12 lie on a line by using slopes.
34. Weight of a Body The weight of a body varies inversely
with the square of its distance from the center of Earth.
Assuming that the radius of Earth is 3960 miles, how much
would a man weigh at an altitude of 1 mile above Earth’s
surface if he weighs 200 pounds on Earth’s surface?
35. Heat Loss The amount of heat transferred per hour through
a glass window varies jointly with the surface area of the
window and the difference in temperature between the
areas separated by the glass. A window with a surface area of
7.5 square feet loses 135 Btu per hour when the temperature
difference is 40°F. How much heat is lost per hour for a
similar window with a surface are of 12 square feet when the
temperature difference is 35°F?
and
33. Mortgage Payments The monthly payment p on a mortgage
varies directly with the amount borrowed B. If the monthly
payment on a 30-year mortgage is $854.00 when $130,000
is borrowed, find an equation that relates the monthly
payment p to the amount borrowed B for a mortgage with
these terms. Then find the monthly payment p when the
amount borrowed B is $165,000.
The Chapter Test Prep Videos are step-by-step solutions available in
, or on this text’s
Channel. Flip back to the Resources
for Success page for a link to this text’s YouTube channel.
Chapter Test
In Problems 1–3, use P1 = 1 - 1, 32 and P2 = 15, - 12 .
1. Find the distance from P1 to P2.
2. Find the midpoint of the line segment joining P1 and P2.
3. (a) Find the slope of the line containing P1 and P2.
(b) Interpret this slope.
4. Graph y = x2 - 9 by plotting points.
5. Sketch the graph of y2 = x.
6. List the intercepts and test for symmetry: x2 + y = 9
7. Write the slope–intercept form of the line with slope - 2
containing the point 13, - 42 . Graph the line.
8. Write the general form of the circle with center 14, - 32 and
radius 5.
9. Find the center and radius of the circle
x2 + y2 + 4x - 2y - 4 = 0.
Graph this circle.
10. For the line 2x + 3y = 6, find a line parallel to it containing
the point 11, - 12 . Also find a line perpendicular to it
containing the point 10, 32 .
11. Resistance Due to a Conductor The resistance (in ohms)
of a circular conductor varies directly with the length of the
conductor and inversely with the square of the radius of the
conductor. If 50 feet of wire with a radius of 6 * 10-3 inch
has a resistance of 10 ohms, what would be the resistance
of 100 feet of the same wire if the radius were increased to
7 * 10-3 inch?
Cumulative Review
In Problems 1–8, find the real solution(s), if any, of each equation.
1. 3x - 5 = 0
2. x - x - 12 = 0
3. 2x - 5x - 3 = 0
4. x2 - 2x - 2 = 0
5. x2 + 2x + 5 = 0
6. 22x + 1 = 3
2
7. 0 x - 2 0 = 1
2
8. 2x2 + 4x = 2
In Problems 9 and 10, solve each equation in the complex
number system.
9. x2 = - 9
10. x2 - 2x + 5 = 0
In Problems 11–14, solve each inequality. Graph the solution set.
11. 2x - 3 … 7
13. 0 x - 2 0 … 1
12. - 1 6 x + 4 6 5
14. 0 2 + x 0 7 3
Chapter Project
15. Find the distance between the points P = 1 - 1, 32 and
Q = 14, - 22 . Find the midpoint of the line segment from
P to Q.
16. Which of the following points are on the graph of
y = x3 - 3x + 1?
(a) 1 - 2, - 12
(b) 12, 32
(c) 13, 12
17. Sketch the graph of y = x3.
197
18. Find the equation of the line containing the points 1 - 1, 42
and 12, - 22 . Express your answer in slope–intercept form.
19. Find the equation of the line perpendicular to the line
y = 2x + 1 and containing the point 13, 52 . Express your
answer in slope–intercept form and graph the line.
20. Graph the equation x2 + y2 - 4x + 8y - 5 = 0.
Chapter Project
The graph below, called a scatter diagram, shows the points
(291.5, 268), (320, 305), c, (368, 385) in a Cartesian plane. From
the graph, it appears that the data follow a linear relation.
Sale Price ($ thousands)
Zestimate vs. Sale Price
in Oak Park, IL
Internet-based Project
Determining the Selling Price of a Home Determining how
much to pay for a home is one of the more difficult decisions
that must be made when purchasing a home. There are many
factors that play a role in a home’s value. Location, size, number of
bedrooms, number of bathrooms, lot size, and building materials
are just a few. Fortunately, the website Zillow.com has developed
its own formula for predicting the selling price of a home. This
information is a great tool for predicting the actual sale price. For
example, the data below show the “zestimate”—the selling price
of a home as predicted by the folks at Zillow—and the actual
selling price of the home, for homes in Oak Park, Illinois.
380
360
340
320
300
280
300
320
340
360
Zestimate ($ thousands)
1. Imagine drawing a line through the data that appears to fit
the data well. Do you believe the slope of the line would be
positive, negative, or close to zero? Why?
2. Pick two points from the scatter diagram. Treat the zestimate
as the value of x, and treat the sale price as the corresponding
value of y. Find the equation of the line through the two
points you selected.
3. Interpret the slope of the line.
4. Use your equation to predict the selling price of a home
whose zestimate is $335,000.
Zestimate
($ thousands)
Sale Price
($ thousands)
291.5
268
320
305
371.5
375
303.5
283
351.5
350
314
275
(b) Select two points from the scatter diagram and find the
equation of the line through the points.
332.5
356
(c) Interpret the slope.
295
300
313
285
368
385
(d) Find a home from the Zillow website that interests
you under the “Make Me Move” option for which a
zestimate is available. Use your equation to predict the
sale price based on the zestimate.
5. Do you believe it would be a good idea to use the equation
you found in part 2 if the zestimate were $950,000? Why or
why not?
6. Choose a location in which you would like to live. Go to
www.zillow.com and randomly select at least ten homes that
have recently sold.
(a) Draw a scatter diagram of your data.
3
Functions and
Their Graphs
Choosing a Wireless Data Plan
Most consumers choose a cellular provider first and then select an appropriate data
plan from that provider. The choice as to the type of plan selected depends on your
use of the device. For example, is online gaming important? Do you want to stream
audio or video? The mathematics learned in this chapter can help you decide what
plan is best suited to your particular needs.
—See the Internet-based Chapter Project—
Outline
3.1
3.2
3.3
3.4
3.5
3.6
198
Functions
The Graph of a Function
Properties of Functions
Library of Functions; Piecewise-defined
Functions
Graphing Techniques: Transformations
Mathematical Models: Building
Functions
Chapter Review
Chapter Test
Cumulative Review
Chapter Projects
A Look Back
So far, our discussion has focused on techniques for graphing equations containing
two variables.
A Look Ahead
In this chapter, we look at a special type of equation involving two variables
called a function. This chapter deals with what a function is, how to graph functions,
properties of functions, and how functions are used in applications. The word
function apparently was introduced by René Descartes in 1637. For him, a function
was simply any positive integral power of a variable x. Gottfried Wilhelm Leibniz
(1646–1716), who always emphasized the geometric side of mathematics, used
the word function to denote any quantity associated with a curve, such as the
coordinates of a point on the curve. Leonhard Euler (1707–1783) employed the
word to mean any equation or formula involving variables and constants. His
idea of a function is similar to the one most often seen in courses that precede
calculus. Later, the use of functions in investigating heat flow equations led to a
very broad definition that originated with Lejeune Dirichlet (1805–1859), which
describes a function as a correspondence between two sets. That is the definition
used in this text.
SECTION 3.1 Functions
199
3.1 Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
r Intervals (Section 1.5, pp. 120–121)
r Solving Inequalities (Section 1.5, pp. 123–126)
r Evaluating Algebraic Expressions, Domain of a
Variable (Chapter R, Section R.2, pp. 20–23)
r Rationalizing Denominators (Chapter R,
Section R.8, p. 75)
Now Work the ‘Are You Prepared?’ problems on page 210.
OBJECTIVES 1
2
3
4
5
1 Determine Whether a Relation Represents a Function
y
5
(1, 2)
24
22
(0, 1)
Determine Whether a Relation Represents a Function (p. 199)
Find the Value of a Function (p. 202)
Find the Difference Quotient of a Function (p. 205)
Find the Domain of a Function Defined by an Equation (p. 206)
Form the Sum, Difference, Product, and Quotient of Two Functions (p. 208)
2
4
x
25
Figure 1 y = 3x - 1
EXAMPL E 1
Often there are situations where the value of one variable is somehow linked to the value of
another variable. For example, an individual’s level of education is linked to annual income.
Engine size is linked to gas mileage. When the value of one variable is related to the value
of a second variable, we have a relation. A relation is a correspondence between two sets.
If x and y are two elements, one from each of these sets, and if a relation exists between
x and y, then we say that x corresponds to y or that y depends on x, and we write x S y.
There are a number of ways to express relations between two sets. For example,
the equation y = 3x - 1 shows a relation between x and y. It says that if we take
some number x, multiply it by 3, and then subtract 1, we obtain the corresponding
value of y. In this sense, x serves as the input to the relation, and y is the output of
the relation. This relation, expressed as a graph, is shown in Figure 1.
The set of all inputs for a relation is called the domain of the relation, and the
set of all outputs is called the range.
In addition to being expressed as equations and graphs, relations can be expressed
through a technique called mapping. A map illustrates a relation as a set of inputs
with an arrow drawn from each element in the set of inputs to the corresponding
element in the set of outputs. Ordered pairs can be used to represent x S y as 1x, y2.
Maps and Ordered Pairs as Relations
Figure 2 shows a relation between states and the number of representatives each
state has in the House of Representatives. (Source: www.house.gov). The relation
might be named “number of representatives.”
State
Number of
Representatives
Alaska
Arizona
California
Colorado
Florida
North Dakota
1
7
9
27
53
Figure 2 Number of representatives
In this relation, Alaska corresponds to 1, Arizona corresponds to 9, and so on. Using
ordered pairs, this relation would be expressed as
5 1Alaska, 12, 1Arizona, 92, 1California, 532, 1Colorado, 72, 1Florida, 272, 1North Dakota, 12 6
200
CHAPTER 3 Functions and Their Graphs
Person
Phone Number
Dan
555 – 2345
Gizmo
549 – 9402
930 – 3956
Colleen
555 – 8294
Phoebe
839 – 9013
Figure 3 Phone numbers
Animal
Life
Expectancy
Dog
11
Duck
10
Kangaroo
Rabbit
7
Figure 4 Animal life expectancy
DEFINITION
y
x
X
Domain
Range
Y
The domain of the relation is {Alaska, Arizona, California, Colorado, Florida,
North Dakota}, and the range is {1, 7, 9, 27, 53}. Note that the output “1” is listed only
once in the range.
r
One of the most important concepts in algebra is the function. A function is a
special type of relation. To understand the idea behind a function, let’s revisit the
relation presented in Example 1. If we were to ask, “How many representatives
does Alaska have?” you would respond “1.” In fact, each input state corresponds to
a single output number of representatives.
Let’s consider a second relation, one that involves a correspondence between
four people and their phone numbers. See Figure 3. Notice that Colleen has two
telephone numbers. There is no single answer to the question “What is Colleen’s
phone number?”
Let’s look at one more relation. Figure 4 is a relation that shows a correspondence
between type of animal and life expectancy. If asked to determine the life
expectancy of a dog, we would all respond, “11 years.” If asked to determine the life
expectancy of a rabbit, we would all respond, “7 years.”
Notice that the relations presented in Figures 2 and 4 have something in
common. What is it? In both of these relations, each input corresponds to exactly
one output. This leads to the definition of a function.
Let X and Y be two nonempty sets.* A function from X into Y is a relation that
associates with each element of X exactly one element of Y.
The set X is called the domain of the function. For each element x in X, the
corresponding element y in Y is called the value of the function at x, or the image
of x. The set of all images of the elements in the domain is called the range of the
function. See Figure 5.
Since there may be some elements in Y that are not the image of some x in X,
it follows that the range of a function may be a subset of Y, as shown in Figure 5.
Not all relations between two sets are functions. The next example shows how
to determine whether a relation is a function.
Figure 5
EX AMPLE 2
Determining Whether a Relation Is a Function
For each relation in Figures 6, 7, and 8, state the domain and range. Then determine
whether the relation is a function.
(a) See Figure 6. For this relation, the input is the number of calories in a fast-food
sandwich, and the output is the fat content (in grams).
Calories
Fat
(Wendy’s 1/4-lb Single) 580
31
(Burger King Whopper) 650
37
(Culver’s Deluxe Single) 541
33
(McDonald's Big Mac) 550
29
(Five Guys Hamburger) 700
43
Figure 6 Fat content
Source: Each company’s Website
*The sets X and Y will usually be sets of real numbers, in which case a (real) function results. The two
sets can also be sets of complex numbers, and then we have defined a complex function. In the broad
definition (proposed by Lejeune Dirichlet), X and Y can be any two sets.
SECTION 3.1 Functions
201
(b) See Figure 7. For this relation, the inputs are gasoline stations in Harris County,
Texas, and the outputs are the price per gallon of unleaded regular in March 2014.
(c) See Figure 8. For this relation, the inputs are the weight (in carats) of pear-cut
diamonds and the outputs are the price (in dollars).
Gas Station
Price per Gallon
Valero
$3.19
Shell
$3.29
Texaco
$3.35
Carats
Price
0.70
$1529
0.71
$1575
0.75
$1765
0.78
$1798
Citgo
$1952
Figure 7 Unleaded price per gallon
Figure 8 Diamond price
Source: Used with permission of Diamonds.com
Solution
(a) The domain of the relation is {541, 550, 580, 650, 700}, and the range of the
relation is {29, 31, 33, 37, 43}. The relation in Figure 6 is a function because each
element in the domain corresponds to exactly one element in the range.
(b) The domain of the relation is {Citgo, Shell, Texaco, Valero}. The range of the
relation is {$3.19, $3.29, $3.35}. The relation in Figure 7 is a function because
each element in the domain corresponds to exactly one element in the range.
Notice that it is okay for more than one element in the domain to correspond
to the same element in the range (Shell and Citgo both sell gas for $3.29 a
gallon).
(c) The domain of the relation is {0.70, 0.71, 0.75, 0.78} and the range is {$1529, $1575,
$1765, $1798, $1952}. The relation in Figure 8 is not a function because not every
element in the domain corresponds to exactly one element in the range. If a
0.71-carat diamond is chosen from the domain, a single price cannot be assigned
to it.
r
Now Work
In Words
For a function, no input has more
than one output. The domain of a
function is the set of all inputs;
the range is the set of all outputs.
EXAMPL E 3
PROBLEM
19
The idea behind a function is its predictability. If the input is known, we can
use the function to determine the output. With “nonfunctions,” we don’t have this
predictability. Look back at Figure 6. If asked, “How many grams of fat are in a
580-calorie sandwich?” we could use the correspondence to answer, “31.” Now consider
Figure 8. If asked, “What is the price of a 0.71-carat diamond?” we could not give
a single response because two outputs result from the single input “0.71.” For this
reason, the relation in Figure 8 is not a function.
We may also think of a function as a set of ordered pairs 1x, y2 in which no
ordered pairs have the same first element and different second elements. The set of
all first elements x is the domain of the function, and the set of all second elements y
is its range. Each element x in the domain corresponds to exactly one element y in
the range.
Determining Whether a Relation Is a Function
For each relation, state the domain and range. Then determine whether the relation
is a function.
(a) 5 11, 42, 12, 52, 13, 62, 14, 72 6
(b) 5 11, 42, 12, 42, 13, 52, 16, 102 6
(c) 5 1 - 3, 92, 1 - 2, 42, 10, 02, 11, 12 , 1 - 3, 82 6
202
CHAPTER 3 Functions and Their Graphs
Solution
(a) The domain of this relation is {1, 2, 3, 4}, and its range is {4, 5, 6, 7}. This
relation is a function because there are no ordered pairs with the same first
element and different second elements.
(b) The domain of this relation is {1, 2, 3, 6}, and its range is {4, 5, 10}. This
relation is a function because there are no ordered pairs with the same first
element and different second elements.
(c) The domain of this relation is { - 3, - 2, 0, 1}, and its range is {0, 1, 4, 8, 9}. This
relation is not a function because there are two ordered pairs, 1 - 3, 92 and
1 - 3, 82, that have the same first element and different second elements.
r
In Example 3(b), notice that 1 and 2 in the domain both have the same image in the
range. This does not violate the definition of a function; two different first elements can
have the same second element. A violation of the definition occurs when two ordered
pairs have the same first element and different second elements, as in Example 3(c).
Now Work
PROBLEM
23
Up to now we have shown how to identify when a relation is a function for
relations defined by mappings (Example 2) and ordered pairs (Example 3). But
relations can also be expressed as equations. The circumstances under which
equations are functions are discussed next.
To determine whether an equation, where y depends on x, is a function, it is often
easiest to solve the equation for y. If any value of x in the domain corresponds to more
than one y, the equation does not define a function; otherwise, it does define a function.
EX AMPLE 4
Determining Whether an Equation Is a Function
Determine whether the equation y = 2x - 5 defines y as a function of x.
Solution
The equation tells us to take an input x, multiply it by 2, and then subtract 5. For any
input x, these operations yield only one output y, so the equation is a function. For
example, if x = 1, then y = 2112 - 5 = - 3. If x = 3, then y = 2132 - 5 = 1.
The graph of the equation y = 2x - 5 is a line with slope 2 and y-intercept - 5. The
function is called a linear function.
r
EX AMPLE 5
Determining Whether an Equation Is a Function
Determine whether the equation x2 + y2 = 1 defines y as a function of x.
Solution
To determine whether the equation x2 + y2 = 1, which defines the unit circle, is a
function, solve the equation for y.
x2 + y 2 = 1
y 2 = 1 - x2
y = { 21 - x2
For values of x for which - 1 6 x 6 1, two values of y result. For example, if x = 0,
then y = {1, so two different outputs result from the same input. This means that
the equation x2 + y2 = 1 does not define a function.
r
Now Work
PROBLEM
37
2 Find the Value of a Function
Functions are often denoted by letters such as f, F, g, G, and others. If f is a function,
then for each number x in its domain, the corresponding image in the range is
designated by the symbol f1x2, read as “ f of x” or as “ f at x.”We refer to f1x2 as the value
of f at the number x ; f1x2 is the number that results when x is given and the function f
is applied; f1x2 is the output corresponding to x or f1x2 is the image of x; f1x2
SECTION 3.1 Functions
203
does not mean “ f times x.” For example, the function given in Example 4 may be
written as y = f1x2 = 2x - 5. Then f112 = - 3 and f132 = 1.
Figure 9 illustrates some other functions. Notice that in every function, for each
x in the domain, there is one value in the range.
1 f (1) f (1)
1
1
0
0 f (0)
2
2 f ( 2)
x
2
1–2 F(2)
1
1 F(1)
f(x ) x 2
Domain
Range
F(4)
1
F(x ) –
x
x
Domain
Range
1
(b) F (x ) –
x
(a) f (x ) x 2
0
0 g(0)
0
1
1 g(1)
2
2 g(2)
2
1–
4
4
3 G(0) G(2) G(3)
3
2 g(4)
4
x
g(x) x
Domain
Range
G(x) 3
x
Domain
Range
(d) G (x) 3
(c) g(x) x
Figure 9
Sometimes it is helpful to think of a function f as a machine that receives as
input a number from the domain, manipulates it, and outputs a value. See Figure 10.
The restrictions on this input/output machine are as follows:
Input x
x
f
1. It accepts only numbers from the domain of the function.
2. For each input, there is exactly one output (which may be repeated for different
inputs).
Output
y f(x)
Figure 10 Input/output machine
EXAMPL E 6
For a function y = f1x2, the variable x is called the independent variable,
because it can be assigned any of the permissible numbers from the domain. The
variable y is called the dependent variable, because its value depends on x.
Any symbols can be used to represent the independent and dependent
variables. For example, if f is the cube function, then f can be given by f1x2 = x3
or f1t2 = t 3 or f1z2 = z3. All three functions are the same. Each says to cube
the independent variable to get the output. In practice, the symbols used for the
independent and dependent variables are based on common usage, such as using C
for cost in business.
The independent variable is also called the argument of the function. Thinking
of the independent variable as an argument can sometimes make it easier to find
the value of a function. For example, if f is the function defined by f1x2 = x3, then
f tells us to cube the argument. Thus f122 means to cube 2, f1a2 means to cube the
number a, and f1x + h2 means to cube the quantity x + h.
Finding Values of a Function
For the function f defined by f1x2 = 2x2 - 3x, evaluate
(a) f132
(e) - f1x2
(b) f1x2 + f 132
(f) f13x2
(c) 3f1x2
(g) f1x + 32
(d) f1 - x2
204
CHAPTER 3 Functions and Their Graphs
Solution
(a) Substitute 3 for x in the equation for f , f(x) = 2x2 - 3x, to get
f 132 = 2132 2 - 3132 = 18 - 9 = 9
The image of 3 is 9.
(b) f1x2 + f 132 = 12x2 - 3x2 + 192 = 2x2 - 3x + 9
(c) Multiply the equation for f by 3.
3f1x2 = 312x2 - 3x2 = 6x2 - 9x
(d) Substitute - x for x in the equation for f and simplify.
f1 - x2 = 21 - x2 2 - 31 - x2 = 2x2 + 3x
Notice the use of parentheses here.
(e) - f1x2 = - 12x2 - 3x2 = - 2x2 + 3x
(f) Substitute 3x for x in the equation for f and simplify.
f 13x2 = 213x2 2 - 313x2 = 219x2 2 - 9x = 18x2 - 9x
(g) Substitute x + 3 for x in the equation for f and simplify.
f1x + 32 = 21x + 32 2 - 31x + 32
= 21x2 + 6x + 92 - 3x - 9
= 2x2 + 12x + 18 - 3x - 9
= 2x2 + 9x + 9
r
Notice in this example that f 1x + 32 ≠ f 1x2 + f132 , f 1 - x2 ≠ - f1x2 , and
3f1x2 ≠ f13x2.
Now Work
PROBLEM
43
Most calculators have special keys that allow you to find the value of certain
commonly used functions. For example, you should be able to find the square
function f1x2 = x2, the square root function f1x2 = 1x, the reciprocal function
1
f1x2 = = x -1, and many others that will be discussed later in this text (such as
x
ln x and log x). Verify the results of Example 7, which follows, on your calculator.
EX AMPLE 7
Finding Values of a Function on a Calculator
(a) f1x2 = x2
1
(b) F1x2 =
x
(c) g1x2 = 1x
f11.2342 = 1.2342 = 1.522756
1
F11.2342 =
≈ 0.8103727715
1.234
g11.2342 = 11.234 ≈ 1.110855526
r
COMMENT Graphing calculators can be used to evaluate any function. Figure 11 shows the result
obtained in Example 6(a) on a TI-84 Plus C graphing calculator with the function to be evaluated,
f (x) = 2x 2 - 3x, in Y1.
Figure 11 Evaluating f (x) = 2x2 - 3x for x = 3
■
SECTION 3.1 Functions
205
Implicit Form of a Function
In general, when a function f is defined by an equation in x and y, we say that the
function f is given implicitly. If it is possible to solve the equation for y in terms of x,
then we write y = f 1x2 and say that the function is given explicitly. For example,
COMMENT The explicit form of a
function is the form required by a
graphing calculator.
■
Implicit Form
3x + y = 5
Explicit Form
y = f 1x2 = - 3x + 5
x2 - y = 6
y = f1x2 = x2 - 6
4
y = f1x2 =
x
xy = 4
SUMMARY
Important Facts about Functions
(a) For each x in the domain of a function f, there is exactly one image f 1x2 in the range; however, an element in
the range can result from more than one x in the domain.
(b) f is the symbol that we use to denote the function. It is symbolic of the equation (rule) that we use to get from
an x in the domain to f1x2 in the range.
(c) If y = f 1x2, then x is called the independent variable or argument of f, and y is called the dependent variable
or the value of f at x.
3 Find the Difference Quotient of a Function
An important concept in calculus involves looking at a certain quotient. For a given
function y = f 1x2, the inputs x and x + h, h ≠ 0, result in the images f1x2 and
f1x + h2. The quotient of their differences
f1x + h2 - f1x2
f1x + h2 - f1x2
=
1x + h2 - x
h
with h ≠ 0, is called the difference quotient of f at x.
DEFINITION
The difference quotient of a function f at x is given by
f1x + h2 - f1x2
h
h ≠ 0
(1)
The difference quotient is used in calculus to define the derivative, which leads to
applications such as the velocity of an object and optimization of resources.
When finding a difference quotient, it is necessary to simplify the expression
in order to cancel the h in the denominator, as illustrated in the following example.
EXAMPL E 8
Finding the Difference Quotient of a Function
Find the difference quotient of each function.
(a) f1x2 = 2x2 - 3x
4
(b) f1x2 =
x
(c) f1x2 = 1x
206
CHAPTER 3 Functions and Their Graphs
Solution
(a)
f1x + h2 - f 1x2
3 21x + h2 2 - 31x + h2 4 - 3 2x2 - 3x4
=
h
h
c
f (x + h) = 2(x + h)2 - 3(x + h)
=
=
=
=
=
(b)
f1x + h2 - f1x2
=
h
=
=
=
=
(c)
21x2 + 2xh + h2 2 - 3x - 3h - 2x2 + 3x
Simplify.
h
2x2 + 4xh + 2h2 - 3h - 2x2
Distribute and combine like terms.
h
4xh + 2h2 - 3h
Combine like terms.
h
h 14x + 2h - 32
Factor out h.
h
4x + 2h - 3
Divide out the h’s.
4
4
x
x + h
4
f (x + h) =
x + h
h
4x - 41x + h2
x1x + h2
Subtract.
h
4x - 4x - 4h
Divide and distribute.
x1x + h2h
- 4h
Simplify.
x1x + h2h
4
Divide out the factor h.
x1x + h2
f1x + h2 - f1x2
2x + h - 2x
=
h
h
=
=
=
=
Now Work
f (x + h) = 1x + h
2x + h - 2x # 2x + h + 2x
h
2x + h + 2x
1 2x + h 2 2 - 1 2x 2 2
h 1 2x + h + 2x 2
h
h 1 2x + h + 2x 2
1
2x + h + 2x
PROBLEM
1 2x
Rationalize the numerator.
(A - B )(A + B ) = A2 - B 2
+ h22 -
1 2x 2 2
= x + h - x = h
Divide out the factor h.
r
79
4 Find the Domain of a Function Defined by an Equation
Often the domain of a function f is not specified; instead, only the equation defining
the function is given. In such cases, we agree that the domain of f is the largest set of
real numbers for which the value f 1x2 is a real number. The domain of a function f
is the same as the domain of the variable x in the expression f1x2.
EX AMPLE 9
Finding the Domain of a Function
Find the domain of each of the following functions.
(a) f1x2 = x2 + 5x
(b) g1x2 =
3x
x - 4
(c) h 1t2 = 24 - 3t
(d) F(x) =
23x + 12
x - 5
2
SECTION 3.1 Functions
Solution
In Words
The domain of g found in
Example 9(b) is
5 x x ≠ - 2, x ≠ 2 6 .
This notation is read, “The domain
of the function g is the set of all
real numbers x such that x does
not equal - 2 and x does not
equal 2.”
207
(a) The function says to square a number and then add five times the number. Since
these operations can be performed on any real number, the domain of f is the
set of all real numbers.
(b) The function g says to divide 3x by x2 - 4. Since division by 0 is not defined, the
denominator x2 - 4 can never be 0, so x can never equal - 2 or 2. The domain
of the function g is 5 x x ≠ - 2, x ≠ 26 .
(c) The function h says to take the square root of 4 - 3t. But only nonnegative
numbers have real square roots, so the expression under the square root (the
radicand) must be nonnegative (greater than or equal to zero). This requires
that
4 - 3t Ú 0
- 3t Ú - 4
4
t …
3
4
4
f , or the interval a - q , d .
3
3
(d) The function F says to take the square root of 3x + 12 and divide this result
by x - 5. This requires that 3x + 12 Ú 0, so x Ú - 4, and also that x - 5 ≠ 0,
so x ≠ 5. Combining these two restrictions, the domain of F is
The domain of h is e t ` t …
5 x x Ú - 4,
x ≠ 56 .
r
The following steps may prove helpful for finding the domain of a function that
is defined by an equation and whose domain is a subset of the real numbers.
Finding the Domain of a Function Defined by an Equation
1. Start with the domain as the set of real numbers.
2. If the equation has a denominator, exclude any numbers that give a zero
denominator.
3. If the equation has a radical of even index, exclude any numbers that cause
the expression inside the radical (the radicand) to be negative.
Now Work
PROBLEM
55
If x is in the domain of a function f, we shall say that f is defined at x, or f(x)
exists. If x is not in the domain of f, we say that f is not defined at x, or f(x) does not
x
exist. For example, if f 1x2 = 2
, then f102 exists, but f112 and f1 - 12 do not
x - 1
exist. (Do you see why?)
We have not said much about finding the range of a function. We will say
more about finding the range when we look at the graph of a function in the next
section. When a function is defined by an equation, it can be difficult to find the
range. Therefore, we shall usually be content to find just the domain of a function
when the function is defined by an equation. We shall express the domain of a
function using inequalities, interval notation, set notation, or words, whichever is
most convenient.
When we use functions in applications, the domain may be restricted by
physical or geometric considerations. For example, the domain of the function f
defined by f 1x2 = x2 is the set of all real numbers. However, if f is used to obtain
the area of a square when the length x of a side is known, then we must restrict
the domain of f to the positive real numbers, since the length of a side can never
be 0 or negative.
208
CHAPTER 3 Functions and Their Graphs
EX A MPL E 1 0
Finding the Domain in an Application
Express the area of a circle as a function of its radius. Find the domain.
Solution
A
r
Figure 12 Circle of radius r
See Figure 12. The formula for the area A of a circle of radius r is A = pr 2. Using
r to represent the independent variable and A to represent the dependent variable,
the function expressing this relationship is
A 1r2 = pr 2
In this setting, the domain is 5 r r 7 06 . (Do you see why?)
r
Observe, in the solution to Example 10, that the symbol A is used in two ways:
It is used to name the function, and it is used to symbolize the dependent variable.
This double use is common in applications and should not cause any difficulty.
Now Work
PROBLEM
97
5 Form the Sum, Difference, Product, and Quotient
of Two Functions
Next we introduce some operations on functions. Functions, like numbers, can
be added, subtracted, multiplied, and divided. For example, if f1x2 = x2 + 9 and
g1x2 = 3x + 5, then
f1x2 + g1x2 = 1x2 + 92 + 13x + 52 = x2 + 3x + 14
The new function y = x2 + 3x + 14 is called the sum function f + g. Similarly,
f1x2 # g1x2 = 1x2 + 92 13x + 52 = 3x3 + 5x2 + 27x + 45
The new function y = 3x3 + 5x2 + 27x + 45 is called the product function f # g.
The general definitions are given next.
DEFINITION
If f and g are functions:
The sum f + g is the function defined by
1f + g2 1x2 = f1x2 + g1x2
In Words
Remember, the symbol x stands
for intersection. It means you
should find the elements that are
common to two sets.
DEFINITION
The domain of f + g consists of the numbers x that are in the domains of both
f and g. That is, domain of f + g = domain of f ∩ domain of g.
The difference f − g is the function defined by
1f - g2 1x2 = f1x2 - g1x2
The domain of f - g consists of the numbers x that are in the domains of both
f and g. That is, domain of f - g = domain of f ∩ domain of g.
DEFINITION
The product f ~ g is the function defined by
1f # g2 1x2 = f1x2 # g1x2
The domain of f # g consists of the numbers x that are in the domains of both f
and g. That is, domain of f # g = domain of f ∩ domain of g.
SECTION 3.1 Functions
DEFINITION
The quotient
209
f
is the function defined by
g
f1x2
f
a b 1x2 =
g
g1x2
g1x2 ≠ 0
f
consists of the numbers x for which g1x2 ≠ 0 and that are in
g
the domains of both f and g. That is,
The domain of
domain of
EX AM PL E 11
f
= {x 0 g(x) ≠ 0} ∩ domain of f ∩ domain of g
g
Operations on Functions
Let f and g be two functions defined as
1
x
and g1x2 =
x + 2
x - 1
Find the following functions, and determine the domain in each case.
f
(a) 1f + g2 1x2
(b) 1f - g2 1x2
(c) 1f # g2 1x2
(d) a b 1x2
g
f1x2 =
Solution
The domain of f is 5 x x ≠ - 26 and the domain of g is 5 x x ≠ 16 .
(a) 1f + g2 1x2 = f 1x2 + g1x2 =
1
x
+
x + 2
x - 1
x1x
+ 22
x - 1
x2 + 3x - 1
=
+
=
1x + 22 1x - 12
1x + 22 1x - 12
1x + 22 1x - 12
The domain of f + g consists of those numbers x that are in the domains of both
f and g. Therefore, the domain of f + g is 5 x x ≠ - 2, x ≠ 16 .
(b) 1f - g2 1x2 = f1x2 - g1x2 =
1
x
x + 2
x - 1
x1x
- 1x2 + x + 12
+ 22
x - 1
=
=
1x + 22 1x - 12
1x + 22 1x - 12
1x + 22 1x - 12
The domain of f - g consists of those numbers x that are in the domains of both
f and g. Therefore, the domain of f - g is 5 x x ≠ - 2, x ≠ 16 .
(c) 1f # g2 1x2 = f 1x2 # g1x2 =
1 # x
x
=
x + 2 x - 1
1x + 22 1x - 12
The domain of f # g consists of those numbers x that are in the domains of both
f and g. Therefore, the domain of f # g is 5 x x ≠ - 2, x ≠ 16 .
1
f 1x2
f
x + 2
1 #x - 1
x - 1
(d) a b 1x2 =
=
=
=
g
x
x
g1x2
x + 2
x1x + 22
x - 1
f
The domain of consists of the numbers x for which g1x2 ≠ 0 and that are in
g
the domains of both f and g. Since g1x2 = 0 when x = 0, we exclude 0 as well
f
as - 2 and 1 from the domain. The domain of is 5 x x ≠ - 2, x ≠ 0, x ≠ 16 .
g
Now Work
PROBLEM
67
r
210
CHAPTER 3 Functions and Their Graphs
In calculus, it is sometimes helpful to view a complicated function as the sum,
difference, product, or quotient of simpler functions. For example,
F1x2 = x2 + 1x is the sum of f1x2 = x2 and g1x2 = 1x.
H1x2 =
x2 - 1
x2 + 1
is the quotient of f1x2 = x2 - 1 and g1x2 = x2 + 1.
SUMMARY
Function
A relation between two sets of real numbers so that each number x in the first set, the
domain, has corresponding to it exactly one number y in the second set.
A set of ordered pairs 1x, y2 or 1x, f1x2 2 in which no first element is paired with two
different second elements.
The range is the set of y-values of the function that are the images of the x-values in the
domain.
A function f may be defined implicitly by an equation involving x and y or explicitly by
writing y = f1x2 .
Unspecified domain
If a function f is defined by an equation and no domain is specified, then the domain will be
taken to be the largest set of real numbers for which the equation defines a real number.
Function notation
y = f1x2
f is a symbol for the function.
x is the independent variable, or argument.
y is the dependent variable.
f1x2 is the value of the function at x, or the image of x.
3.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. The inequality - 1 6 x 6 3 can be written in interval
notation as
. (pp. 120–121)
2. If x = - 2, the value of the
1
.(pp. 20–23)
expression 3x2 - 5x + is
x
x - 3
is
3. The domain of the variable in the expression
x + 4
. (pp. 20–23)
4. Solve the inequality: 3 - 2x 7 5. Graph the solution
set. (pp. 123–126)
3
, multiply the
5. To rationalize the denominator of
25 - 2
numerator and denominator by
(p. 75)
6. A quotient is considered rationalized if its denominator
. (p. 75)
contains no
Concepts and Vocabulary
7. If f is a function defined by the equation y = f 1x2, then x
is called the
variable, and y is the
variable.
8. If the domain of f is all real numbers in the interval 30, 74 ,
and the domain of g is all real numbers in the interval
3 - 2, 54, then the domain of f + g is all real numbers in the
interval
.
f
9. The domain of consists of numbers x for which g1x2
0
g
and
.
that are in the domains of both
10. If f 1x2 = x + 1 and g1x2 = x3,
then
= x3 - 1x + 12.
11. True or False Every relation is a function.
12. True or False The domain of 1f # g2 1x2 consists of the
numbers x that are in the domains of both f and g.
13. True or False If no domain is specified for a function f, then
the domain of f is taken to be the set of real numbers.
14. True or False The domain of the function f 1x2 =
is 5x x ≠ {26 .
x2 - 4
x
15. The set of all images of the elements in the domain of a
function is called the
.
(a) range (b) domain (c) solution set (d) function
16. The independent variable is sometimes referred to as the
of the function.
(a) range (b) value (c) argument (d) definition
f(x + h) - f(x)
is called the
of f.
17. The expression
h
(a) radicand
(b) image
(c) correspondence
(d) difference quotient
18. When written as y = f(x), a function is said to be
defined
.
(a) explicitly
(b) consistently
(c) implicitly
(d) rationally
SECTION 3.1 Functions
211
Skill Building
In Problems 19–30, state the domain and range for each relation. Then determine whether each relation represents a function.
19.
Person
20.
Birthday
Elvis
Daughter
Father
Jan. 8
Bob
Kaleigh
Mar. 15
John
Linda
Marissa
Sept. 17
Chuck
Marcia
Beth
Colleen
Diane
21. Hours Worked
22.
Salary
20 Hours
$200
$300
30 Hours
$350
40 Hours
$425
23. 5 12, 62, 1 - 3, 62, 14, 92, 12, 102 6
26. 5 10, - 22, 11, 32, 12, 32, 13, 72 6
29. 5 1 - 2, 42, 1 - 1, 12, 10, 02, 11, 12 6
Level of Education
Average Income
Less than 9th grade
9th-12th grade
High School Graduate
Some College
College Graduate
$18,120
$23,251
$36,055
$45,810
$67,165
24. 5 1 - 2, 52, 1 - 1, 32, 13, 72, 14, 122 6 25. 5 11, 32, 12, 32, 13, 32, 14, 32 6
27. 5 1 - 2, 42, 1 - 2, 62, 10, 32, 13, 72 6
28. 5 1 - 4, 42, 1 - 3, 32, 1 - 2, 22, 1 - 1, 12, 1 - 4, 02 6
30. 5 1 - 2, 162, 1 - 1, 42, 10, 32, 11, 42 6
In Problems 31–42, determine whether the equation defines y as a function of x.
34. y = 0 x 0
1
x
31. y = 2x2 - 3x + 4
32. y = x3
33. y =
35. y2 = 4 - x2
36. y = { 21 - 2x
37. x = y2
38. x + y2 = 1
3
39. y = 2
x
40. y =
41. 2x2 + 3y2 = 1
42. x2 - 4y2 = 1
3x - 1
x + 2
In Problems 43–50, find the following for each function:
(a) f 102 (b) f 112 (c) f 1 - 12 (d) f 1 - x2 (e) - f 1x2
(f) f 1x + 12
(g) f 12x2
(h) f 1x + h2
x2 - 1
x + 4
43. f 1x2 = 3x2 + 2x - 4
44. f 1x2 = - 2x2 + x - 1
x
45. f 1x2 = 2
x + 1
47. f 1x2 = 0 x 0 + 4
48. f 1x2 = 2x2 + x
49. f 1x2 =
2x + 1
3x - 5
52. f 1x2 = x2 + 2
53. f 1x2 =
x
x + 1
54. f 1x2 =
2x
x2 - 4
57. F 1x2 =
x - 2
x3 + x
58. G1x2 =
2
Ax - 1
62. f 1x2 =
46. f 1x2 =
50. f 1x2 = 1 -
1
1x + 22 2
In Problems 51–66, find the domain of each function.
51. f 1x2 = - 5x + 4
55. g1x2 =
x
x2 - 16
56. h1x2 =
2
59. h1x2 = 23x - 12
60. G1x2 = 21 - x
61. p1x2 =
63. f 1x2 =
64. f 1x2 =
65. P(t) =
x
2x - 4
-x
2- x - 2
2t - 4
3t - 21
66. h(z) =
In Problems 67–76, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.
f
(b) 1f - g2 1x2
(c) 1f # g2 1x2
(d) a b 1x2
(a) 1f + g2 1x2
g
f
(e) 1f + g2 132
(f) 1f - g2 142
(g) 1f # g2 122
(h) a b 112
g
67. f 1x2 = 3x + 4; g1x2 = 2x - 3
69. f 1x2 = x - 1; g1x2 = 2x
2
68. f 1x2 = 2x + 1; g1x2 = 3x - 2
70. f 1x2 = 2x2 + 3; g1x2 = 4x3 + 1
x2
x + 1
2
x + 4
x3 - 4x
4
2x - 9
2z + 3
z - 2
212
CHAPTER 3 Functions and Their Graphs
71. f 1x2 = 1x; g1x2 = 3x - 5
72. f 1x2 = 0 x 0 ; g1x2 = x
73. f 1x2 = 1 +
74. f 1x2 = 2x - 1; g1x2 = 24 - x
1
1
; g1x2 =
x
x
2x + 3
4x
75. f 1x2 =
; g1x2 =
3x - 2
3x - 2
76. f 1x2 = 2x + 1; g1x2 =
77. Given f 1x2 = 3x + 1 and 1f + g2 1x2 = 6 -
1
x, find the
2
78. Given f 1x2 =
2
x
f
x + 1
1
, find the function g.
and a b 1x2 = 2
x
g
x - x
function g.
In Problems 79–90, find the difference quotient of f; that is, find
f 1x + h2 - f 1x2
h
, h ≠ 0, for each function. Be sure to simplify.
79. f 1x2 = 4x + 3
80. f 1x2 = - 3x + 1
81. f(x) = x2 - 4
83. f 1x2 = x2 - x + 4
84. f 1x2 = 3x2 - 2x + 6
85. f(x) =
87. f(x) =
2x
x + 3
88. f(x) =
5x
x - 4
82. f(x) = 3x2 + 2
86. f 1x2 =
1
x2
89. f(x) = 2x - 2
1
x + 3
90. f(x) = 2x + 1
[Hint: Rationalize the
numerator.]
Applications and Extensions
91. Given f(x) = x2 - 2x + 3, find the value(s) for x such that
f(x) = 11.
3
5
92. Given f(x) = x - , find the value(s) for x such that
6
4
7
f(x) = - .
16
93. If f 1x2 = 2x + Ax + 4x - 5 and f 122 = 5, what is the
value of A ?
3
2
94. If f 1x2 = 3x2 - Bx + 4 and f 1 - 12 = 12, what is the value
of B ?
3x + 8
95. If f 1x2 =
and f 102 = 2, what is the value of A ?
2x - A
96. If f 1x2 =
2x - B
1
and f 122 = , what is the value of B ?
3x + 4
2
represents the number N of housing units (in millions) in
2012 that had r rooms, where r is an integer and 2 … r … 9.
(a) Identify the dependent and independent variables.
(b) Evaluate N(3). Provide a verbal explanation of the
meaning of N(3).
103. Effect of Gravity on Earth If a rock falls from a height of
20 meters on Earth, the height H (in meters) after x seconds
is approximately
H1x2 = 20 - 4.9x2
(a) What is the height of the rock when x = 1 second?
x = 1.1 seconds? x = 1.2 seconds? x = 1.3 seconds?
(b) When is the height of the rock 15 meters? When is it
10 meters? When is it 5 meters?
(c) When does the rock strike the ground?
97. Geometry Express the area A of a rectangle as a function
of the length x if the length of the rectangle is twice its width.
104. Effect of Gravity on Jupiter If a rock falls from a height of
20 meters on the planet Jupiter, its height H (in meters) after
x seconds is approximately
98. Geometry Express the area A of an isosceles right triangle
as a function of the length x of one of the two equal sides.
H1x2 = 20 - 13x2
99. Constructing Functions Express the gross salary G of a
person who earns $14 per hour as a function of the number x
of hours worked.
100. Constructing Functions Tiffany, a commissioned salesperson,
earns $100 base pay plus $10 per item sold. Express her
gross salary G as a function of the number x of items sold.
(a) What is the height of the rock when x = 1 second?
x = 1.1 seconds? x = 1.2 seconds?
(b) When is the height of the rock 15 meters? When is it
10 meters? When is it 5 meters?
(c) When does the rock strike the ground?
101. Population as a Function of Age The function
P 1a2 = 0.014a2 - 5.073a + 327.287
represents the population P (in millions) of Americans that
are a years of age or older in 2012.
(a) Identify the dependent and independent variables.
(b) Evaluate P 1202 . Provide a verbal explanation of the
meaning of P 1202 .
(c) Evaluate P 102 . Provide a verbal explanation of the
meaning of P 102 .
Source: U.S. Census Bureau
102. Number of Rooms The function
N1r2 = - 1.35r 2 + 15.45r - 20.71
105. Cost of Trans-Atlantic Travel A Boeing 747 crosses the
Atlantic Ocean (3000 miles) with an airspeed of 500 miles
per hour. The cost C (in dollars) per passenger is given by
C 1x2 = 100 +
36,000
x
+
10
x
SECTION 3.1 Functions
where x is the ground speed 1airspeed { wind2.
(a) What is the cost per passenger for quiescent (no wind)
conditions?
(b) What is the cost per passenger with a head wind of
50 miles per hour?
(c) What is the cost per passenger with a tail wind of
100 miles per hour?
(d) What is the cost per passenger with a head wind of
100 miles per hour?
106. Cross-sectional Area The cross-sectional area of a beam
cut from a log with radius 1 foot is given by the function
A1x2 = 4x11 - x2 , where x represents the length, in feet,
of half the base of the beam. See the figure. Determine the
cross-sectional area of the beam if the length of half the base
of the beam is as follows:
(a) One-third of a foot
A(x ) 4x 1 x 2
(b) One-half of a foot
(c) Two-thirds of a foot
1
x
107. Economics The participation rate is the number of
people in the labor force divided by the civilian population
(excludes military). Let L 1x2 represent the size of the labor
force in year x, and P 1x2 represent the civilian population in
year x. Determine a function that represents the participation
rate R as a function of x.
108. Crimes Suppose that V 1x2 represents the number of violent
crimes committed in year x and P 1x2 represents the number of
property crimes committed in year x. Determine a function T
that represents the combined total of violent crimes and
property crimes in year x.
109. Health Care Suppose that P 1x2 represents the percentage
of income spent on health care in year x and I 1x2 represents
income in year x. Determine a function H that represents
total health care expenditures in year x.
213
110. Income Tax Suppose that I 1x2 represents the income of an
individual in year x before taxes and T 1x2 represents the
individual’s tax bill in year x. Determine a function N that
represents the individual’s net income (income after taxes)
in year x.
111. Profit Function Suppose that the revenue R, in dollars, from
selling x cell phones, in hundreds, is R 1x2 = - 1.2x2 + 220x.
The cost C, in dollars, of selling x cell phones, in hundreds, is
C 1x2 = 0.05x3 - 2x2 + 65x + 500.
(a) Find the profit function, P 1x2 = R 1x2 - C 1x2.
(b) Find the profit if x = 15 hundred cell phones are sold.
(c) Interpret P(15).
112. Profit Function Suppose that the revenue R, in dollars,
from selling x clocks is R1x2 = 30x. The cost C, in dollars, of
selling x clocks is C 1x2 = 0.1x2 + 7x + 400.
(a) Find the profit function, P 1x2 = R 1x2 - C 1x2.
(b) Find the profit if x = 30 clocks are sold.
(c) Interpret P(30).
113. Stopping Distance When the driver of a vehicle observes an
impediment, the total stopping distance involves both the
reaction distance (the distance the vehicle travels while
the driver moves his or her foot to the brake pedal) and
the braking distance (the distance the vehicle travels once the
brakes are applied). For a car traveling at a speed of
v miles per hour, the reaction distance R, in feet, can be
estimated by R(v) = 2.2v. Suppose that the braking distance
B, in feet, for a car is given by B(v) = 0.05v2 + 0.4v - 15.
(a) Find the stopping distance function
D(v) = R(v) + B(v).
(b) Find the stopping distance if the car is traveling at a
speed of 60 mph.
(c) Interpret D(60).
114. Some functions f have the property that f 1a + b2 =
f 1a2 + f 1b2 for all real numbers a and b. Which of the
following functions have this property?
(a) h1x2 = 2x
(b) g1x2 = x2
1
(c) F 1x2 = 5x - 2 (d) G1x2 =
x
Explaining Concepts: Discussion and Writing
x2 - 1
the
115. Are the functions f 1x2 = x - 1 and g1x2 =
x + 1
same? Explain.
117. Find a function H that multiplies a number x by 3 and then
subtracts the cube of x and divides the result by your age.
116. Investigate when, historically, the use of the function
notation y = f 1x2 first appeared.
Retain Your Knowledge
Problems 118–121 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
118. List
the
intercepts
and
test
for
symmetry:
(x + 12)2 + y2 = 16
119. Determine which of the given points are on the graph of
the equation. y = 3x2 - 82x
120. How many pounds of lean hamburger that is 7% fat must
be mixed with 12 pounds of ground chuck that is 20% fat
to have a hamburger mixture that is 15% fat?
121. Solve x3 - 9x = 2x2 - 18.
Points: ( - 1, - 5), (4, 32), (9, 171)
‘Are You Prepared?’ Answers
1. 1 - 1, 32
2. 21.5
3. 5 x x ≠ - 46
4. 5x x 6 - 16
1
0
1
5. 25 + 2
6. radicals
214
CHAPTER 3 Functions and Their Graphs
3.2 The Graph of a Function
PREPARING FOR THIS SECTION Before getting started, review the following:
r Graphs of Equations (Section 2.2, pp. 157–159)
r Intercepts (Section 2.2, pp. 159–160)
Now Work the ‘Are You Prepared?’ problems on page 218.
OBJECTIVES 1 Identify the Graph of a Function (p. 214)
2 Obtain Information from or about the Graph of a Function (p. 215)
In applications, a graph often demonstrates more clearly the relationship between
two variables than, say, an equation or table. For example, Table 1 shows the average
price of gasoline in the United States for the years 1985–2014 (adjusted for inflation,
based on 2014 dollars). If we plot these data and then connect the points, we obtain
Figure 13.
1.80
2006
3.01
1987
1.89
1997
1.76
2007
3.19
1988
1.81
1998
1.49
2008
3.56
1989
1.87
1999
1.61
2009
2.58
1990
2.03
2000
2.03
2010
3.00
1991
1.91
2001
1.90
2011
3.69
1992
1.82
2002
1.76
2012
3.72
1993
1.74
2003
1.99
2013
3.54
1994
1.71
2004
2.31
2014
3.43
2013
2014
1996
2011
1.90
2009
1986
4.00
3.50
3.00
2.50
2.00
1.50
1.00
0.50
0.00
2007
2.74
2005
2005
2003
1.72
2001
1995
1999
2.55
1997
1985
1995
Price
1993
Year
1991
Price
1989
Year
1987
Price
1985
Year
Price (dollars per gallon)
Table 1
Figure 13 Average retail price of gasoline (2014 dollars)
Source: U.S. Energy Information Administration
Source: U.S. Energy Information Administration
We can see from the graph that the price of gasoline (adjusted for inflation)
stayed roughly the same from 1986 to 1991 and rose rapidly from 2002 to 2008. The
graph also shows that the lowest price occurred in 1998. To learn information such
as this from an equation requires that some calculations be made.
Look again at Figure 13. The graph shows that for each date on the horizontal
axis, there is only one price on the vertical axis. The graph represents a function,
although the exact rule for getting from date to price is not given.
When a function is defined by an equation in x and y, the graph of the function
is the graph of the equation; that is, it is the set of points 1x, y2 in the xy-plane that
satisfy the equation.
1 Identify the Graph of a Function
In Words
If any vertical line intersects a
graph at more than one point,
the graph is not the graph of a
function.
THEOREM
Not every collection of points in the xy-plane represents the graph of a function.
Remember, for a function, each number x in the domain has exactly one image y
in the range. This means that the graph of a function cannot contain two points
with the same x-coordinate and different y-coordinates. Therefore, the graph of a
function must satisfy the following vertical-line test.
Vertical-Line Test
A set of points in the xy-plane is the graph of a function if and only if every
vertical line intersects the graph in at most one point.
SECTION 3.2 The Graph of a Function
215
Identifying the Graph of a Function
EXAMPL E 1
Which of the graphs in Figure 14 are graphs of functions?
y
6
y
4
y
y
3
1
(1, 1)
4
3
(a) y x 2
Figure 14
Solution
3x
4x
4
(1, 1)
6 x
1
1
3
(b) y x 3
1 x
(c) x y 2
(d) x 2 y 2 1
The graphs in Figures 14(a) and 14(b) are graphs of functions, because every vertical
line intersects each graph in at most one point. The graphs in Figures 14(c) and 14(d)
are not graphs of functions, because there is a vertical line that intersects each graph
in more than one point. Notice in Figure 14(c) that the input 1 corresponds to two
outputs, - 1 and 1. This is why the graph does not represent a function.
r
Now Work
PROBLEM
17
2 Obtain Information from or about the Graph of a Function
If 1x, y2 is a point on the graph of a function f, then y is the value of f at x; that is,
y = f1x2 . Also if y = f1x2, then 1x, y2 is a point on the graph of f. For example, if
1 - 2, 72 is on the graph of f, then f 1 - 22 = 7, and if f 152 = 8, then the point 15, 82
is on the graph of y = f1x2 . The next example illustrates how to obtain information
about a function if its graph is given.
Obtaining Information from the Graph of a Function
EXAMPL E 2
Let f be the function whose graph is given in Figure 15. (The graph of f might
represent the distance y that the bob of a pendulum is from its at-rest position at time x.
Negative values of y mean that the pendulum is to the left of the at-rest position, and
positive values of y mean that the pendulum is to the right of the at-rest position.)
y
4
2
(5––2, 0)
(––2 , 0)
(7––2, 0)
(3––2, 0)
2
4
(4, 4)
(2, 4)
(0, 4)
(, 4)
x
(3, 4)
Figure 15
(a) What are f102, f a
(b)
(c)
(d)
(e)
(f)
(g)
Solution
3p
b , and f13p2?
2
What is the domain of f ?
What is the range of f ?
List the intercepts. (Recall that these are the points, if any, where the graph
crosses or touches the coordinate axes.)
How many times does the line y = 2 intersect the graph?
For what values of x does f1x2 = - 4?
For what values of x is f1x2 7 0?
(a) Since 10, 42 is on the graph of f, the y-coordinate 4 is the value of f at the
3p
x-coordinate 0; that is, f102 = 4. In a similar way, when x =
, then y = 0, so
2
3p
f a b = 0. When x = 3p, then y = - 4, so f 13p2 = - 4.
2
(b) To determine the domain of f, notice that the points on the graph of f have
x-coordinates between 0 and 4p, inclusive; and for each number x between 0 and
4p, there is a point 1x, f1x2 2 on the graph. The domain of f is 5 x 0 … x … 4p6
or the interval 3 0, 4p4 .
(c) The points on the graph all have y-coordinates between - 4 and 4, inclusive; and
for each such number y, there is at least one number x in the domain. The range
of f is 5 y - 4 … y … 46 or the interval 3 - 4, 44 .
216
CHAPTER 3 Functions and Their Graphs
(d) The intercepts are the points
p
3p
5p
10, 42, a , 0b , a , 0b , a , 0b , and
2
2
2
a
7p
, 0b
2
(e) Draw the horizontal line y = 2 on the graph in Figure 15. Notice that the line
intersects the graph four times.
(f) Since 1p, - 42 and 13p, - 42 are the only points on the graph for which
y = f1x2 = - 4, we have f1x2 = - 4 when x = p and x = 3p.
(g) To determine where f1x2 7 0, look at Figure 15 and determine the
x-values from 0 to 4p for which the y-coordinate is positive. This occurs
p
3p 5p
7p
b h a ,
b h a , 4p d . Using inequality notation, f1x2 7 0
2
2 2
2
p
3p
5p
7p
for 0 … x 6
or
6 x 6
or
6 x … 4p.
2
2
2
2
on c 0,
r
When the graph of a function is given, its domain may be viewed as the shadow
created by the graph on the x-axis by vertical beams of light. Its range can be viewed
as the shadow created by the graph on the y-axis by horizontal beams of light. Try
this technique with the graph given in Figure 15.
Now Work
EX AMPLE 3
PROBLEMS
11
AND
15
Obtaining Information about the Graph of a Function
Consider the function: f1x2 =
x + 1
x + 2
(a) Find the domain of f.
1
(b) Is the point a1, b on the graph of f ?
2
(c) If x = 2, what is f1x2? What point is on the graph of f ?
(d) If f1x2 = 2, what is x? What point is on the graph of f ?
(e) What are the x-intercepts of the graph of f (if any)? What point(s) are on the
graph of f ?
Solution
(a) The domain of f is 5 x x ≠ - 26 .
(b) When x = 1, then
f112 =
2
1 + 1
=
1 + 2
3
f1x2 =
x + 1
x + 2
1
2
The point a1, b is on the graph of f; the point a1, b is not.
3
2
(c) If x = 2, then
f122 =
2 + 1
3
=
2 + 2
4
3
The point ¢ 2, ≤ is on the graph of f.
4
(d) If f1x2 = 2, then
x
x
x
x
+
+
+
+
1
2
1
1
x
= 2
f1x2 = 2
= 21x + 22
= 2x + 4
= -3
Multiply both sides by x + 2.
Distribute.
Solve for x.
If f1x2 = 2, then x = - 3. The point 1 - 3, 22 is on the graph of f.
SECTION 3.2 The Graph of a Function
217
(e) The x-intercepts of the graph of f are the real solutions of the equation f1x2 = 0
that are in the domain of f.
x + 1
= 0
x + 2
x + 1 = 0
x = -1
Multiply both sides by x + 2.
Subtract 1 from both sides.
x + 1
= 0 is x = - 1, so - 1 is
x + 2
the only x-intercept. Since f1 - 12 = 0, the point ( - 1, 0) is on the graph of f.
The only real solution of the equation f1x2 =
r
Now Work
EXAMPL E 4
PROBLEM
27
Average Cost Function
The average cost C per computer of manufacturing x computers per day is given by
the function
C 1x2 = 0.56x2 - 34.39x + 1212.57 +
20,000
x
Determine the average cost of manufacturing:
(a) 30 computers in a day
(b) 40 computers in a day
(c) 50 computers in a day
(d) Graph the function C = C 1x2, 0 6 x … 80.
(e) Create a TABLE with TblStart = 1 and ∆Tbl = 1. Which value of x minimizes
the average cost?
Solution
(a) The average cost per computer of manufacturing x = 30 computers is
C 1302 = 0.561302 2 - 34.391302 + 1212.57 +
20,000
= $1351.54
30
(b) The average cost per computer of manufacturing x = 40 computers is
C 1402 = 0.561402 2 - 34.391402 + 1212.57 +
20,000
= $1232.97
40
(c) The average cost per computer of manufacturing x = 50 computers is
C 1502 = 0.561502 2 - 34.391502 + 1212.57 +
(d) See Figure 16 for the graph of C = C 1x2.
20,000
= $1293.07
50
(e) With the function C = C 1x2 in Y1 , we create Table 2. We scroll down until
we find a value of x for which Y1 is smallest. Table 3 shows that manufacturing
x = 41 computers minimizes the average cost at $1231.74 per computer.
4000
0
0
80
Figure 16 C1x2 = 0.56x2 - 34.39x + 1212.57 +
20,000
x
Now Work
Table 2
PROBLEM
Table 3
35
r
218
CHAPTER 3 Functions and Their Graphs
SUMMARY
Graph of a Function The collection of points 1x, y2 that satisfies the equation y = f 1x2.
Vertical-Line Test
A collection of points is the graph of a function if and only if every vertical line intersects
the graph in at most one point.
3.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. The intercepts of the equation x2 + 4y2 = 16 are
(pp. 159–160)
.
2. True or False The point 1 - 2, - 62 is on the graph of the
equation x = 2y - 2. (pp. 157–159)
Concepts and Vocabulary
3. A set of points in the xy-plane is the graph of a function if
and only if every
line intersects the graph in at
most one point.
4. If the point 15, - 32 is a point on the graph of f, then
f1
2 =
.
5. Find a so that the point 1 - 1, 22 is on the graph of
f 1x2 = ax2 + 4.
6. True or False Every graph represents a function.
7. True or False The graph of a function y = f 1x2 always
crosses the y-axis.
8. True or False The y-intercept of the graph of the function
y = f 1x2, whose domain is all real numbers, is f 102.
9. If a function is defined by an equation in x and y, then the
set of points (x, y) in the xy-plane that satisfy the equation is
called
.
(a) the domain of the function (b) the range of the function
(c) the graph of the function (d) the relation of the function
10. The graph of a function y = f(x) can have more than one of
which type of intercept?
(a) x-intercept (b) y-intercept (c) both (d) neither
Skill Building
11. Use the given graph of the function f to answer parts (a)–(n).
y
(0, 3) 4
12. Use the given graph of the function f to answer parts (a)–(n).
y
(2, 4)
(4, 3)
4
(10, 0) (11, 1)
(5, 3)
(–4, 2)
(–2, 1) 2
(–3, 0)
(4, 0)
5
–5
(6, 0)
(–5, –2)
(–6, –3)
–3
11 x
(8, – 2)
Find f 102 and f 1 - 62.
Find f 162 and f 1112.
Is f 132 positive or negative?
Is f 1 - 42 positive or negative?
For what values of x is f 1x2 = 0?
For what values of x is f 1x2 7 0?
What is the domain of f ?
What is the range of f ?
What are the x-intercepts?
What is the y-intercept?
1
(k) How often does the line y = intersect the graph?
2
(l) How often does the line x = 5 intersect the graph?
(m) For what values of x does f 1x2 = 3?
(n) For what values of x does f 1x2 = - 2?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
–4
–2
(0, 0) –2
2
4
(6, 0)
6
x
(2, –2)
(a) Find f 102 and f 162.
(b) Find f 122 and f 1 - 22.
(c) Is f 132 positive or negative?
(d) Is f 1 - 12 positive or negative?
(e) For what values of x is f 1x2 = 0?
(f) For what values of x is f 1x2 6 0?
(g) What is the domain of f ?
(h) What is the range of f ?
(i) What are the x-intercepts?
(j) What is the y-intercept?
(k) How often does the line y = - 1 intersect the graph?
(l) How often does the line x = 1 intersect the graph?
(m) For what value of x does f 1x2 = 3?
(n) For what value of x does f 1x2 = - 2?
SECTION 3.2 The Graph of a Function
219
In Problems 13–24, determine whether the graph is that of a function by using the vertical-line test. If it is, use the graph to find:
(a) The domain and range
(b) The intercepts, if any
(c) Any symmetry with respect to the x-axis, the y-axis, or the origin
13.
14.
y
3
3
3
3x
3
y
(1, 2) 3
22.
3
3
23.
y
3
3
26. f 1x2 = - 3x2 + 5x
(a) Is the point 1 - 1, 22 on the graph of f ?
(b) If x = - 2, what is f 1x2? What point is on the graph of f ?
(c) If f 1x2 = - 2, what is x? What point(s) are on the graph
of f ?
(d) What is the domain of f ?
(e) List the x-intercepts, if any, of the graph of f.
(f) List the y-intercept, if there is one, of the graph of f.
x + 2
x - 6
Is the point 13, 142 on the graph of f ?
If x = 4, what is f 1x2? What point is on the graph of f ?
If f 1x2 = 2, what is x? What point(s) are on the graph
of f ?
What is the domain of f ?
List the x-intercepts, if any, of the graph of f.
List the y-intercept, if there is one, of the graph of f.
28. f 1x2 =
x2 + 2
x + 4
20.
y
4
(3, 2)
3
(a) Is the point a1, b on the graph of f ?
5
x
(4, 3)
4
4x
4
24.
y
3x
27. f 1x2 =
––
2
1
y
(1–2 , 5)
4
6
3
––
2
9
25. f 1x2 = 2x2 - x - 1
(a) Is the point 1 - 1, 22 on the graph of f ?
(b) If x = - 2, what is f 1x2? What point is on the graph of f ?
(c) If f 1x2 = - 1, what is x? What point(s) are on the graph
of f ?
(d) What is the domain of f ?
(e) List the x-intercepts, if any, of the graph of f.
(f) List the y-intercept, if there is one, of the graph of f.
(d)
(e)
(f)
3
In Problems 25–30, answer the questions about the given function.
(a)
(b)
(c)
x
3x
x
3
––
2
3
3x
(1, 2)
––
2
1
1
y
3
3
3
21.
19.
y
3
3
3x
y
1
3x
18.
y
3
16.
y
3
3
17.
15.
y
3
3
1
3
3 x
(2, 3)
3
3 x
(b) If x = 0, what is f 1x2? What point is on the graph of f ?
1
(c) If f 1x2 = , what is x? What point(s) are on the graph
2
of f ?
(d) What is the domain of f ?
(e) List the x-intercepts, if any, of the graph of f.
(f) List the y-intercept, if there is one, of the graph of f.
2x2
x4 + 1
Is the point 1 - 1, 12 on the graph of f ?
If x = 2, what is f 1x2? What point is on the graph of f ?
If f 1x2 = 1, what is x? What point(s) are on the graph
of f ?
What is the domain of f ?
List the x-intercepts, if any, of the graph of f.
List the y-intercept, if there is one, of the graph of f.
29. f 1x2 =
(a)
(b)
(c)
(d)
(e)
(f)
30. f 1x2 =
2x
x - 2
2
1
(a) Is the point a , - b on the graph of f ?
2
3
(b) If x = 4, what is f 1x2? What point is on the graph of f ?
(c) If f 1x2 = 1, what is x? What point(s) are on the graph
of f ?
(d) What is the domain of f ?
(e) List the x-intercepts, if any, of the graph of f.
(f) List the y-intercept, if there is one, of the graph of f.
220
CHAPTER 3 Functions and Their Graphs
Applications and Extensions
31. The graphs of two functions, f and g, are illustrated. Use the
graphs to answer parts (a)–(f).
y
yg(x)
(2, 2)
2 (2, 1)
(4, 1)
2
2
(6, 1)
(6, 0)
x
yf(x)
4
(3, 2)
(5, 2)
(4, 3)
4
(a) 1f + g2 122
(c) 1f - g2 162
(e) 1f # g2 122
(b) 1f + g2 142
(d) 1g - f2 162
f
(f) a b 142
g
32. Granny Shots The last player in the NBA to use an underhand
foul shot (a “granny” shot) was Hall of Fame forward Rick
Barry, who retired in 1980. Barry believes that current NBA
players could increase their free-throw percentage if they
were to use an underhand shot. Since underhand shots are
released from a lower position, the angle of the shot must
be increased. If a player shoots an underhand foul shot,
releasing the ball at a 70-degree angle from a position 3.5 feet
above the floor, then the path of the ball can be modeled
136x2
by the function h1x2 = + 2.7x + 3.5, where h is
v2
the height of the ball above the floor, x is the forward distance
of the ball in front of the foul line, and v is the initial velocity
with which the ball is shot in feet per second.
(a) The center of the hoop is 10 feet above the floor and
15 feet in front of the foul line. Determine the initial
velocity with which the ball must be shot in order for the
ball to go through the hoop.
(b) Write the function for the path of the ball using the velocity
found in part (a).
(c) Determine the height of the ball after it has traveled
9 feet in front of the foul line.
(d) Find additional points and graph the path of the
basketball.
Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10,
October 2000
33. Free-throw Shots According to physicist Peter Brancazio,
the key to a successful foul shot in basketball lies in the
arc of the shot. Brancazio determined the optimal angle of
the arc from the free-throw line to be 45 degrees. The arc
also depends on the velocity with which the ball is shot. If a
player shoots a foul shot, releasing the ball at a 45-degree
angle from a position 6 feet above the floor, then the path of
the ball can be modeled by the function
h1x2 = -
44x2
+ x + 6
v2
where h is the height of the ball above the floor, x is the
forward distance of the ball in front of the foul line, and v
is the initial velocity with which the ball is shot in feet per
second. Suppose a player shoots a ball with an initial
velocity of 28 feet per second.
(a) Determine the height of the ball after it has traveled
8 feet in front of the foul line.
(b) Determine the height of the ball after it has traveled
12 feet in front of the foul line.
(c) Find additional points and graph the path of the
basketball.
(d) The center of the hoop is 10 feet above the floor and
15 feet in front of the foul line. Will the ball go through
the hoop? Why or why not? If not, with what initial
velocity must the ball be shot in order for the ball to go
through the hoop?
Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10,
October 2000
34. Cross-sectional Area The cross-sectional area of a beam
cut from a log with radius 1 foot is given by the function
A1x2 = 4x21 - x2 , where x represents the length, in
feet, of half the base of the beam. See the figure.
(a) Find the domain of A.
(b) Use a graphing utility to graph the function A = A1x2.
(c) Create a TABLE with TblStart = 0 and ∆Tbl = 0.1
for 0 … x … 1. Which value of x maximizes the crosssectional area? What should be the length of the base of
the beam to maximize the cross-sectional area?
A(x ) 4x 1 x 2
1
x
35. Motion of a Golf Ball A golf ball is hit with an initial
velocity of 130 feet per second at an inclination of 45° to the
horizontal. In physics, it is established that the height h of
the golf ball is given by the function
- 32x2
+ x
1302
where x is the horizontal distance that the golf ball has
traveled.
h1x2 =
SECTION 3.2 The Graph of a Function
2
4000
W 1h2 = ma
b
4000 + h
(a) If Amy weighs 120 pounds at sea level, how much will
she weigh on Pikes Peak, which is 14,110 feet above sea
level?
(b) Use a graphing utility to graph the function W = W 1h2.
Use m = 120 pounds.
(c) Create a TABLE with TblStart = 0 and ∆Tbl = 0.5 to
see how the weight W varies as h changes from 0 to 5 miles.
(d) At what height will Amy weigh 119.95 pounds?
(e) Does your answer to part (d) seem reasonable?
Explain.
37. Cost of Trans-Atlantic Travel A Boeing 747 crosses the
Atlantic Ocean (3000 miles) with an airspeed of 500 miles
per hour. The cost C (in dollars) per passenger is given by
C 1x2 = 100 +
36,000
x
+
10
x
where x is the groundspeed 1airspeed { wind2.
(a) What is the cost when the groundspeed is 480 miles per
hour? 600 miles per hour?
(b) Find the domain of C.
(c) Use a graphing utility to graph the function C = C 1x2.
(d) Create a TABLE with TblStart = 0 and ∆Tbl = 50.
(e) To the nearest 50 miles per hour, what groundspeed
minimizes the cost per passenger?
38. Reading and Interpreting Graphs Let C be the function
whose graph is given in the next column. This graph
represents the cost C of manufacturing q computers in a day.
(a) Determine C(0). Interpret this value.
(b) Determine C(10). Interpret this value.
(c) Determine C(50). Interpret this value.
C
(100, 280 000)
250,000
200,000
150,000
100,000
50,000
(50, 51 000)
(30, 32 000)
(10, 19 000)
(0, 5000)
10
20
30
40 50 60 70 80
Number of computers
90 100 q
39. Reading and Interpreting Graphs Let C be the function
whose graph is given below. This graph represents the cost C
of using g gigabytes of data in a month for a shared-data
family plan.
(a) Determine C (0). Interpret this value.
(b) Determine C (5). Interpret this value.
(c) Determine C (15). Interpret this value.
(d) What is the domain of C? What does this domain imply
in terms of the number of gigabytes?
(e) Describe the shape of the graph.
C
480
(60, 405)
Cost (dollars)
36. Effect of Elevation on Weight If an object weighs m pounds
at sea level, then its weight W (in pounds) at a height of
h miles above sea level is given approximately by
(d) What is the domain of C? What does this domain imply
in terms of daily production?
(e) Describe the shape of the graph.
(f) The point (30, 32 000) is called an inflection point.
Describe the behavior of the graph around the
inflection point.
Cost (dollars per day)
(a) Determine the height of the golf ball after it has
traveled 100 feet.
(b) What is the height after it has traveled 300 feet?
(c) What is the height after it has traveled 500 feet?
(d) How far was the golf ball hit?
(e) Use a graphing utility to graph the function h = h1x2.
(f) Use a graphing utility to determine the distance that
the ball has traveled when the height of the ball is
90 feet.
(g) Create a TABLE with TblStart = 0 and ∆Tbl = 25. To
the nearest 25 feet, how far does the ball travel before
it reaches a maximum height? What is the maximum
height?
(h) Adjust the value of ∆Tbl until you determine the
distance, to within 1 foot, that the ball travels before it
reaches its maximum height.
221
320
160
(5, 30)
(15, 90)
(0, 30)
0
20
40
Gigabytes
60 g
Explaining Concepts: Discussion and Writing
40. Describe how you would find the domain and range of a function if you were given its graph. How would your strategy change if
you were given the equation defining the function instead of its graph?
41. How many x-intercepts can the graph of a function have? How many y-intercepts can the graph of a function have?
42. Is a graph that consists of a single point the graph of a function? Can you write the equation of such a function?
222
CHAPTER 3 Functions and Their Graphs
43. Match each of the following functions with the graph that best describes the situation.
(a) The cost of building a house as a function of its square footage
(b) The height of an egg dropped from a 300-foot building as a function of time
(c) The height of a human as a function of time
(d) The demand for Big Macs as a function of price
(e) The height of a child on a swing as a function of time
y
y
y
y
x
x
x
x
x
(III)
(II)
(I)
y
(V)
(IV)
44. Match each of the following functions with the graph that best describes the situation.
(a) The temperature of a bowl of soup as a function of time
(b) The number of hours of daylight per day over a 2-year period
(c) The population of Florida as a function of time
(d) The distance traveled by a car going at a constant velocity as a function of time
(e) The height of a golf ball hit with a 7-iron as a function of time
y
y
y
x
x
y
x
(II)
(I)
y
x
45. Consider the following scenario: Barbara decides to take
a walk. She leaves home, walks 2 blocks in 5 minutes at a
constant speed, and realizes that she forgot to lock the door.
So Barbara runs home in 1 minute. While at her doorstep, it
takes her 1 minute to find her keys and lock the door. Barbara
walks 5 blocks in 15 minutes and then decides to jog home. It
takes her 7 minutes to get home. Draw a graph of Barbara’s
distance from home (in blocks) as a function of time.
46. Consider the following scenario: Jayne enjoys riding her
bicycle through the woods. At the forest preserve, she gets on her
bicycle and rides up a 2000-foot incline in 10 minutes. She then
travels down the incline in 3 minutes. The next 5000 feet is level
terrain, and she covers the distance in 20 minutes. She rests for
15 minutes. Jayne then travels 10,000 feet in 30 minutes. Draw a
graph of Jayne’s distance traveled (in feet) as a function of time.
47. The following sketch represents the distance d (in miles)
that Kevin was from home as a function of time t (in hours).
Answer the questions by referring to the graph. In parts
(a)–(g), how many hours elapsed and how far was Kevin
from home during this time?
x
(V)
(IV)
(III)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
From t = 0 to t = 2
From t = 2 to t = 2.5
From t = 2.5 to t = 2.8
From t = 2.8 to t = 3
From t = 3 to t = 3.9
From t = 3.9 to t = 4.2
From t = 4.2 to t = 5.3
What is the farthest distance that Kevin was from home?
How many times did Kevin return home?
48. The following sketch represents the speed v (in miles per
hour) of Michael’s car as a function of time t (in minutes).
v (t )
(7, 50)
(2, 30)
(8, 38)
(4, 30)
(4.2, 0)
(7.4, 50)
(7.6, 38)
(6, 0)
(9.1, 0)
t
d (t )
(2, 3)
(2.5, 3)
(2.8, 0)
(3.9, 2.8)
(3, 0)
(4.2, 2.8)
(5.3, 0)
t
(a) Over what interval of time was Michael traveling
fastest?
(b) Over what interval(s) of time was Michael’s speed zero?
(c) What was Michael’s speed between 0 and 2 minutes?
(d) What was Michael’s speed between 4.2 and 6 minutes?
(e) What was Michael’s speed between 7 and 7.4 minutes?
(f) When was Michael’s speed constant?
SECTION 3.3 Properties of Functions
49. Draw the graph of a function whose domain is
5x - 3 … x … 8, x ≠ 56
and
whose
range
is
5y - 1 … y … 2, y ≠ 06. What point(s) in the
rectangle - 3 … x … 8, - 1 … y … 2 cannot be on the
graph? Compare your graph with those of other students.
What differences do you see?
223
50. Is there a function whose graph is symmetric with respect to
the x-axis? Explain.
51. Explain why the vertical-line test works.
Retain Your Knowledge
Problems 52–55 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
52. Factor completely: 8x2 + 24x + 18
53. Find the distance between the points (3, - 6) and (1, 0).
2
54. Write the equation of the line with slope that passes through the point ( - 6, 4).
3
55. Subtract: (4x3 - 5x2 + 2) - (3x2 + 5x - 2)
‘Are You Prepared?’ Answers
1. 1 - 4, 02, 14, 02, 10, - 22, 10, 22
2. False
3.3 Properties of Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
r Intervals (Section 1.5, pp. 120–121)
r Intercepts (Section 2.2, pp. 159–160)
r Slope of a Line (Section 2.3, pp. 167–169)
r Point–Slope Form of a Line (Section 2.3, p. 171)
r Symmetry (Section 2.2, pp. 160–162)
Now Work the ‘Are You Prepared?’ problems on page 232.
OBJECTIVES 1 Determine Even and Odd Functions from a Graph (p. 223)
2 Identify Even and Odd Functions from an Equation (p. 225)
3 Use a Graph to Determine Where a Function Is Increasing, Decreasing,
or Constant (p. 225)
4 Use a Graph to Locate Local Maxima and Local Minima (p. 226)
5 Use a Graph to Locate the Absolute Maximum and the Absolute
Minimum (p. 227)
6 Use a Graphing Utility to Approximate Local Maxima and Local Minima
and to Determine Where a Function Is Increasing or Decreasing (p. 229)
7 Find the Average Rate of Change of a Function (p. 230)
To obtain the graph of a function y = f1x2 , it is often helpful to know certain
properties that the function has and the impact of these properties on the way the
graph will look.
1 Determine Even and Odd Functions from a Graph
The words even and odd, when applied to a function f, describe the symmetry that
exists for the graph of the function.
224
CHAPTER 3 Functions and Their Graphs
A function f is even if and only if, whenever the point 1x, y2 is on the graph
of f , the point 1 - x, y2 is also on the graph. Using function notation, we define an
even function as follows:
DEFINITION
A function f is even if, for every number x in its domain, the number - x is also
in the domain and
f1 - x2 = f1x2
A function f is odd if and only if, whenever the point 1x, y2 is on the graph of f ,
the point 1 - x, - y2 is also on the graph. Using function notation, we define an odd
function as follows:
DEFINITION
A function f is odd if, for every number x in its domain, the number - x is also
in the domain and
f1 - x2 = - f1x2
Refer to page 162, where the tests for symmetry are listed. The following results
are then evident.
THEOREM
EX AMPLE 1
A function is even if and only if its graph is symmetric with respect to the
y-axis. A function is odd if and only if its graph is symmetric with respect to
the origin.
Determining Even and Odd Functions from the Graph
Determine whether each graph given in Figure 17 is the graph of an even function,
an odd function, or a function that is neither even nor odd.
y
y
x
Figure 17
Solution
y
x
(b)
(a)
x
(c)
(a) The graph in Figure 17(a) is that of an even function, because the graph is
symmetric with respect to the y-axis.
(b) The function whose graph is given in Figure 17(b) is neither even nor odd,
because the graph is neither symmetric with respect to the y-axis nor symmetric
with respect to the origin.
(c) The function whose graph is given in Figure 17(c) is odd, because its graph is
symmetric with respect to the origin.
r
Now Work
PROBLEMS
25(a), (b),
AND
(d)
SECTION 3.3 Properties of Functions
225
2 Identify Even and Odd Functions from an Equation
EXAMPL E 2
Identifying Even and Odd Functions Algebraically
Determine whether each of the following functions is even, odd, or neither. Then
determine whether the graph is symmetric with respect to the y-axis, with respect to
the origin, or neither.
(a) f1x2 = x2 - 5
(c) h 1x2 = 5x3 - x
Solution
(b) g1x2 = x3 - 1
(d) F1x2 = 0 x 0
(a) To determine whether f is even, odd, or neither, replace x by - x in f1x2 = x2 - 5.
f 1 - x2 = 1 - x2 2 - 5 = x2 - 5 = f1x2
Since f 1 - x2 = f1x2 , the function is even, and the graph of f is symmetric with
respect to the y-axis.
(b) Replace x by - x in g1x2 = x3 - 1.
g1 - x2 = 1 - x2 3 - 1 = - x3 - 1
Since g1 - x2 ≠ g1x2 and g1- x2 ≠ - g1x2 = - 1x3 - 12 = - x3 + 1, the
function is neither even nor odd. The graph of g is not symmetric with respect
to the y-axis, nor is it symmetric with respect to the origin.
(c) Replace x by - x in h 1x2 = 5x3 - x.
h 1 - x2 = 51 - x2 3 - 1 - x2 = - 5x3 + x = - 15x3 - x2 = - h 1x2
Since h 1 - x2 = - h 1x2, h is an odd function, and the graph of h is symmetric
with respect to the origin.
(d) Replace x by - x in F1x2 = 0 x 0 .
F1 - x2 = 0 - x 0 = 0 - 1 0 # 0 x 0 = 0 x 0 = F1x2
Since F1 - x2 = F1x2, F is an even function, and the graph of F is symmetric
with respect to the y-axis.
r
Now Work
PROBLEM
37
3 Use a Graph to Determine Where a Function Is Increasing,
Decreasing, or Constant
Consider the graph given in Figure 18. If you look from left to right along the graph
of the function, you will notice that parts of the graph are going up, parts are going
down, and parts are horizontal. In such cases, the function is described as increasing,
decreasing, or constant, respectively.
y
5
(0, 4)
(6, 0)
(2, 0)
4
Figure 18
EXAMPL E 3
(4, 2)
(3, 4)
y = f (x )
(6, 1)
6 x
2
Determining Where a Function Is Increasing, Decreasing,
or Constant from Its Graph
Determine the values of x for which the function in Figure 18 is increasing. Where is
it decreasing? Where is it constant?
226
CHAPTER 3 Functions and Their Graphs
Solution
When determining where a function is increasing, where it is decreasing, and where
it is constant, we use strict inequalities involving the independent variable x, or we
use open intervals* of x-coordinates. The function whose graph is given in Figure 18
is increasing on the open interval 1 - 4, 02 , or for - 4 6 x 6 0. The function is
decreasing on the open intervals 1 - 6, - 42 and 13, 62 , or for - 6 6 x 6 - 4 and
3 6 x 6 6. The function is constant on the open interval 10, 32 , or for 0 6 x 6 3.
WARNING Describe the behavior of a
graph in terms of its x-values. Do not say
the graph in Figure 18 is increasing from
the point 1 - 4, - 22 to the point (0, 4).
Rather, say it is increasing on the interval
■
1 - 4, 02 .
DEFINITIONS A function f is increasing on an open interval I if, for any
choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 6 f1x2 2.
In Words
If a function is decreasing, then
as the values of x get bigger, the
values of the function get smaller.
If a function is increasing, then as
the values of x get bigger, the values
of the function also get bigger. If a
function is constant, then as the
values of x get bigger, the values of
the function remain unchanged.
A function f is decreasing on an open interval I if, for any choice of x1 and x2
in I, with x1 6 x2 , we have f1x1 2 7 f 1x2 2.
A function f is constant on an open interval I if, for all choices of x in I, the
values f1x2 are equal.
Figure 19 illustrates the definitions. The graph of an increasing function goes up
from left to right, the graph of a decreasing function goes down from left to right,
and the graph of a constant function remains at a fixed height.
y
y
y
f (x 1)
f (x 2)
f (x 1)
x1
x2
x
f (x 1)
f (x 2)
x1
x
x2
x1
(b) For x 1 < x 2 in l,
f(x 1) > f (x 2);
f is decreasing on I.
(a) For x 1 < x 2 in l,
f (x 1) < f (x 2);
f is increasing on I.
Now Work
PROBLEMS
f(x 2)
x2
x
l
l
l
Figure 19
r
More precise definitions follow:
(c) For all x in I, the values of
f are equal; f is constant on I.
13, 15, 17,
AND
25(c)
4 Use a Graph to Locate Local Maxima and Local Minima
Suppose f is a function defined on an open interval I containing c. If the value of
f at c is greater than or equal to the values of f on I, then f has a local maximum
at c.† See Figure 20(a).
If the value of f at c is less than or equal to the values of f on I, then f has a
local minimum at c. See Figure 20(b).
y
y
f has a local maximum f(c)
at c.
(c, f(c))
c
(a)
f(c)
x
f has a local minimum
at c.
(c, f(c))
c
(b)
Figure 20 Local maximum and local minimum
*The open interval 1a, b2 consists of all real numbers x for which a 6 x 6 b.
†
Some texts use the term relative instead of local.
x
SECTION 3.3 Properties of Functions
DEFINITIONS
227
Let f be a function defined on some interval I.
A function f has a local maximum at c if there is an open interval in I containing c
so that, for all x in this open interval, we have f1x2 … f1c2. We call f1c2 a
local maximum value of f.
A function f has a local minimum at c if there is an open interval in I containing c
so that, for all x in this open interval, we have f 1x2 Ú f1c2. We call f1c2 a
local minimum value of f.
If f has a local maximum at c, then the value of f at c is greater than or equal to
the values of f near c. If f has a local minimum at c, then the value of f at c is less than
or equal to the values of f near c. The word local is used to suggest that it is only near c,
not necessarily over the entire domain, that the value f1c2 has these properties.
EXAMPL E 4
y
y f(x)
Figure 21 shows the graph of a function f.
(1, 2)
2
(–1, 1)
–2
3
Finding Local Maxima and Local Minima from the Graph
of a Function and Determining Where the Function
Is Increasing, Decreasing, or Constant
x
Figure 21
Solution
WARNING The y-value is the local
maximum value or local minimum value,
and it occurs at some x-value. For
example, in Figure 21, we say f has a local
maximum at 1 and the local maximum
value is 2.
■
(a) At what value(s) of x, if any, does f have a local maximum? List the local maximum
values.
(b) At what value(s) of x, if any, does f have a local minimum? List the local minimum
values.
(c) Find the intervals on which f is increasing. Find the intervals on which f is
decreasing.
The domain of f is the set of real numbers.
(a) f has a local maximum at 1, since for all x close to 1, we have f 1x2 … f112 . The
local maximum value is f112 = 2.
(b) f has local minima at - 1 and at 3. The local minimum values are f1 - 12 = 1
and f132 = 0.
(c) The function whose graph is given in Figure 21 is increasing for all values of x
between - 1 and 1 and for all values of x greater than 3. That is, the function is
increasing on the intervals 1 - 1, 12 and 13, q 2 , or for - 1 6 x 6 1 and x 7 3.
The function is decreasing for all values of x less than - 1 and for all values of x
between 1 and 3. That is, the function is decreasing on the intervals 1 - q , - 12
and 11, 32 , or for x 6 - 1 and 1 6 x 6 3.
r
Now Work
y
y f (x)
(b, f (b))
(v, f (v))
a
u
v b
domain: [a, b]
for all x in [a, b], f(x) f(u)
for all x in [a, b], f(x) f(v)
absolute maximum: f(u)
absolute minimum: f(v)
Figure 22
19
AND
21
5 Use a Graph to Locate the Absolute Maximum
and the Absolute Minimum
(u, f(u))
(a, f(a))
PROBLEMS
Look at the graph of the function f given in Figure 22. The domain of f is the closed
interval 3 a, b4 . Also, the largest value of f is f 1u2 and the smallest value of f is f1v2.
These are called, respectively, the absolute maximum and the absolute minimum
of f on 3 a, b 4 .
x
DEFINITION Let f be a function defined on some interval I. If there is
a number u in I for which f 1x2 … f1u2 for all x in I, then f has an absolute
maximum at u, and the number f1u2 is the absolute maximum of f on I.
If there is a number v in I for which f1x2 Ú f 1v2 for all x in I, then f has an
absolute minimum at v, and the number f1v2 is the absolute minimum of f on I.
228
CHAPTER 3 Functions and Their Graphs
The absolute maximum and absolute minimum of a function f are sometimes
called the extreme values of f on I.
The absolute maximum or absolute minimum of a function f may not exist. Let’s
look at some examples.
Finding the Absolute Maximum and the Absolute Minimum
from the Graph of a Function
EX AMPLE 5
For each graph of a function y = f 1x2 in Figure 23, find the absolute maximum and
the absolute minimum, if they exist. Also, find any local maxima or local minima.
(3, 6)
6
4
6
6
(5, 5)
4
(4, 4)
2
2
1
3
5
x
1
Figure 23
3
(b)
Solution
WARNING A function may have an
absolute maximum or an absolute
minimum at an endpoint but not a
local maximum or a local minimum.
Why? Local maxima and local
minima are found over some open interval I,
and this interval cannot be created
around an endpoint.
■
4
4
(1, 4)
x
2
2
(1, 1) (2, 1)
5
6
(4, 3)
2
(3, 1)
(a)
6
(0, 3)
(1, 2)
(0, 1)
(5, 4)
4
(5, 3)
y
y
y
y
y
1
3
(c)
(2, 2)
(0, 0)
5
x
1
3
(d)
5
x
1
3
5
(e)
(a) The function f whose graph is given in Figure 23(a) has the closed interval [0, 5]
as its domain. The largest value of f is f132 = 6, the absolute maximum. The
smallest value of f is f102 = 1, the absolute minimum. The function has a local
maximum of 6 at x = 3 and a local minimum of 4 at x = 4.
(b) The function f whose graph is given in Figure 23(b) has the domain
5 x|1 … x … 5, x ≠ 36 . Note that we exclude 3 from the domain because of the
“hole” at (3, 1). The largest value of f on its domain is f 152 = 3, the absolute
maximum. There is no absolute minimum. Do you see why? As you trace the
graph, getting closer to the point (3, 1), there is no single smallest value. [As soon
as you claim a smallest value, we can trace closer to (3, 1) and get a smaller value!]
The function has no local maxima or minima.
(c) The function f whose graph is given in Figure 23(c) has the interval [0, 5] as
its domain. The absolute maximum of f is f152 = 4. The absolute minimum is 1.
Notice that the absolute minimum 1 occurs at any number in the interval [1, 2].
The function has a local minimum value of 1 at every x in the interval [1, 2], but
it has no local maximum value.
(d) The function f given in Figure 23(d) has the interval 3 0, q ) as its domain. The
function has no absolute maximum; the absolute minimum is f102 = 0. The
function has no local maximum or local minimum.
(e) The function f in Figure 23(e) has the domain 5 x|1 6 x 6 5, x ≠ 26. The function
has no absolute maximum and no absolute minimum. Do you see why? The function
has a local maximum value of 3 at x = 4, but no local minimum value.
r
In calculus, there is a theorem with conditions that guarantee a function will
have an absolute maximum and an absolute minimum.
THEOREM
Extreme Value Theorem
If f is a continuous function* whose domain is a closed interval 3 a, b 4 , then f
has an absolute maximum and an absolute minimum on 3 a, b4 .
*Although a precise definition requires calculus, we’ll agree for now that a continuous function is one
whose graph has no gaps or holes and can be traced without lifting the pencil from the paper.
x
SECTION 3.3 Properties of Functions
229
The absolute maximum (minimum) can be found by selecting the largest (smallest)
value of f from the following list:
1. The values of f at any local maxima or local minima of f in [a, b].
2. The value of f at each endpoint of [a, b]—that is, f(a) and f(b).
For example, the graph of the function f given in Figure 23(a) is continuous on
the closed interval [0, 5]. The Extreme Value Theorem guarantees that f has extreme
values on [0, 5]. To find them, we list
1. The value of f at the local extrema: f 132 = 6, f142 = 4
2. The value of f at the endpoints: f 102 = 1, f152 = 5
The largest of these, 6, is the absolute maximum; the smallest of these, 1, is the
absolute minimum.
Now Work
PROBLEM
49
6 Use a Graphing Utility to Approximate Local Maxima
and Local Minima and to Determine Where a Function
Is Increasing or Decreasing
To locate the exact value at which a function f has a local maximum or a local
minimum usually requires calculus. However, a graphing utility may be used to
approximate these values using the MAXIMUM and MINIMUM features.
EXAM PL E 6
Using a Graphing Utility to Approximate Local Maxima
and Minima and to Determine Where a Function Is Increasing
or Decreasing
(a) Use a graphing utility to graph f1x2 = 6x3 - 12x + 5 for - 2 6 x 6 2.
Approximate where f has a local maximum and where f has a local minimum.
(b) Determine where f is increasing and where it is decreasing.
Solution
(a) Graphing utilities have a feature that finds the maximum or minimum point
of a graph within a given interval. Graph the function f for - 2 6 x 6 2. The
MAXIMUM and MINIMUM commands require us to first determine the
open interval I. The graphing utility will then approximate the maximum or
minimum value in the interval. Using MAXIMUM, we find that the local maximum
value is 11.53 and that it occurs at x = - 0.82, rounded to two decimal places. See
Figure 24(a). Using MINIMUM, we find that the local minimum value is - 1.53
and that it occurs at x = 0.82, rounded to two decimal places. See Figure 24(b).
30
30
−2
−2
2
−10
−10
Figure 24
(a) Local maximum
2
(b) Local minimum
(b) Looking at Figures 24(a) and (b), we see that the graph of f is increasing from
x = - 2 to x = - 0.82 and from x = 0.82 to x = 2, so f is increasing on the
intervals 1 - 2, - 0.822 and 10.82, 22 , or for - 2 6 x 6 - 0.82 and 0.82 6 x 6 2.
The graph is decreasing from x = - 0.82 to x = 0.82, so f is decreasing on the
interval 1 - 0.82, 0.822 , or for - 0.82 6 x 6 0.82.
r
Now Work
PROBLEM
57
230
CHAPTER 3 Functions and Their Graphs
7 Find the Average Rate of Change of a Function
In Section 2.3, we said that the slope of a line can be interpreted as the average rate
of change. To find the average rate of change of a function between any two points
on its graph, calculate the slope of the line containing the two points.
If a and b, a ≠ b, are in the domain of a function y = f 1x2, the average rate
of change of f from a to b is defined as
DEFINITION
In Words
Average rate of change =
The symbol ∆ is the Greek capital
letter delta and is read “change in.”
f1b2 - f1a2
∆y
=
∆x
b - a
a ≠ b
(1)
The symbol ∆y in equation (1) is the “change in y,” and ∆x is the “change in x.”
The average rate of change of f is the change in y divided by the change in x.
Finding the Average Rate of Change
EX AMPLE 7
Find the average rate of change of f1x2 = 3x2:
(a) From 1 to 3
(b) From 1 to 5
(c) From 1 to 7
(a) The average rate of change of f1x2 = 3x2 from 1 to 3 is
Solution
f132 - f112
∆y
27 - 3
24
=
=
=
= 12
∆x
3 - 1
3 - 1
2
(b) The average rate of change of f 1x2 = 3x2 from 1 to 5 is
y
f152 - f112
∆y
75 - 3
72
=
=
=
= 18
∆x
5 - 1
5 - 1
4
160
(7, 147)
120
Average rate
of change 5 24
80
(c) The average rate of change of f 1x2 = 3x2 from 1 to 7 is
(5, 75)
Average rate
of change 5 18
40
(3, 27)
(1, 3)
(0, 0)
2
4
Figure 25 f 1x2 = 3x2
Average rate
of change 5 12
6
x
f172 - f112
∆y
147 - 3
144
=
=
=
= 24
∆x
7 - 1
7 - 1
6
r
See Figure 25 for a graph of f 1x2 = 3x2. The function f is increasing for x 7 0.
The fact that the average rate of change is positive for any x1, x2, x1 ≠ x2, in the
interval 11, 72 indicates that the graph is increasing on 1 6 x 6 7. Further, the
average rate of change is consistently getting larger for 1 6 x 6 7, which indicates
that the graph is increasing at an increasing rate.
Now Work
PROBLEM
65
The Secant Line
The average rate of change of a function has an important geometric interpretation.
Look at the graph of y = f 1x2 in Figure 26. Two points are labeled on the graph:
1a, f1a2 2 and 1b, f1b2 2. The line containing these two points is called the secant
line; its slope is
msec =
f 1b2 - f 1a2
b - a
SECTION 3.3 Properties of Functions
y
231
y 5 f (x )
Secant line
(b, f (b ))
I(b ) 2 f (a )
(a, f (a ))
b2a
b
a
x
Figure 26 Secant line
THEOREM
Slope of the Secant Line
The average rate of change of a function from a to b equals the slope of the
secant line containing the two points 1 a, f 1a2 2 and 1b, f 1b2 2 on its graph.
EXAMPL E 8
Finding the Equation of a Secant Line
Suppose that g1x2 = 3x2 - 2x + 3.
(a) Find the average rate of change of g from - 2 to 1.
(b) Find an equation of the secant line containing 1 - 2, g1 - 22 2 and 11, g112 2 .
(c) Using a graphing utility, draw the graph of g and the secant line obtained in
part (b) on the same screen.
Solution
(a) The average rate of change of g1x2 = 3x2 - 2x + 3 from - 2 to 1 is
Average rate of change =
g112 - g1 - 22
1 - 1 - 22
4 - 19
3
15
= = -5
3
=
g (1) = 3(1)2 - 2(1) + 3 = 4
g ( - 2) = 3( - 2)2 - 2( - 2) + 3 = 19
(b) The slope of the secant line containing 1 - 2, g1 - 22 2 = 1 - 2, 192 and
11, g112 2 = 11, 42 is msec = - 5. Use the point–slope form to find an
equation of the secant line.
24
−3
y - y1 = msec 1x - x1 2
3
−4
Figure 27 Graph of g and the secant line
Point–slope form of the secant line
y - 19 = - 51x - 1 - 22 2
x1 = - 2, y1 = g ( - 2) = 19, msec = - 5
y - 19 = - 5x - 10
Distribute.
y = - 5x + 9
Slope–intercept form of the secant line
(c) Figure 27 shows the graph of g along with the secant line y = - 5x + 9.
Now Work
PROBLEM
71
r
232
CHAPTER 3 Functions and Their Graphs
3.3 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. The interval 12, 52 can be written as the inequality
(pp. 120–121)
.
2. The slope of the line containing the points 1 - 2, 32 and
13, 82 is
. (pp. 167–169)
4. Write the point–slope form of the line with slope 5
containing the point 13, - 22. (p. 171)
5. The intercepts of the equation y = x2 - 9 are
(pp. 159–160)
.
3. Test the equation y = 5x2 - 1 for symmetry with respect to
the x-axis, the y-axis, and the origin. (pp. 160–162)
Concepts and Vocabulary
6. A function f is
on an open interval I if, for any
choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2.
function f is one for which f 1 - x2 = f 1x2 for
7. A(n)
every x in the domain of f; a(n)
function f is one for
which f 1 - x2 = - f 1x2 for every x in the domain of f.
8. True or False A function f is decreasing on an open interval
I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have
f 1x1 2 7 f 1x2 2.
9. True or False A function f has a local maximum at c if
there is an open interval I containing c such that for all x in
I, f 1x2 … f 1c2.
10. True or False Even functions have graphs that are
symmetric with respect to the origin.
11. An odd function is symmetric with respect to
(a) the x-axis (b) the y-axis
(c) the origin (d) the line y = x
.
12. Which of the following intervals is required to guarantee a
continuous function will have both an absolute maximum
and an absolute minimum?
(a) (a, b)
(b) (a, b4
(d) 3a, b4
(c) 3a, b)
Skill Building
In Problems 13–24, use the graph of the function f given.
y
13. Is f increasing on the interval 1 - 8, - 22?
(2, 10)
10
14. Is f decreasing on the interval 1 - 8, - 42?
(22, 6)
15. Is f increasing on the interval 1 - 2, 62?
(210, 0)
16. Is f decreasing on the interval 12, 52?
(5, 0)
(25, 0)
210
17. List the interval(s) on which f is increasing.
18. List the interval(s) on which f is decreasing.
25
(0, 0)
(28, 24)
19. Is there a local maximum at 2? If yes, what is it?
(7, 3)
10 x
5
26
20. Is there a local maximum at 5? If yes, what is it?
21. List the number(s) at which f has a local maximum. What are the local maximum values?
22. List the number(s) at which f has a local minimum. What are the local minimum values?
23. Find the absolute minimum of f on 3 - 10, 74 .
24. Find the absolute maximum of f on 3 - 10, 74 .
In Problems 25–32, the graph of a function is given. Use the graph to find:
(a) The intercepts, if any
(b) The domain and range
(c) The intervals on which the function is increasing, decreasing, or constant
(d) Whether the function is even, odd, or neither
25.
26.
y
4
(4, 2)
(3, 3)
27.
y
3
(0, 3)
28.
y
3
(3, 3)
y
3
(0, 2)
(4, 2)
(0, 1)
4 (2, 0)
(2, 0)
4x
3
(1, 0)
(1, 0)
3 x
3
3 x
3
(1, 0)
3x
233
SECTION 3.3 Properties of Functions
29.
30.
y
2
31.
y
2
(––2 , 1)
( – 3, 2)
(0, 1)
32.
y
3
1
0, –
2
( )
(3, 1)
( –1, 2)
––
2
––
2
x
2
(, 1)
( ––2 , 1)
2
x
1
–, 0
3
( )
2
2
(2, 2)
(2, 1)
(2.3, 0)
(3, 0)
(0, 1)
3
3 x
(2,
–1)
(1, –1)
–3
(, 1)
y
3
3 x
(3, 2)
–3
2
In Problems 33–36, the graph of a function f is given. Use the graph to find:
(a) The numbers, if any, at which f has a local maximum. What are the local maximum values?
(b) The numbers, if any, at which f has a local minimum. What are the local minimum values?
33.
34.
y
4
35.
y
36.
y
(
3
(0, 3)
1
(0, 2)
4 (2, 0)
4x
(2, 0)
3 (1, 0)
3 x
(1, 0)
––
2
––
,
2
1)
––
2
π
x
π2
π
2
π x
(π, 1)
1
( ––2 , 1)
y
2
(π, 1)
2
In Problems 37–48, determine algebraically whether each function is even, odd, or neither.
37. f 1x2 = 4x3
38. f 1x2 = 2x4 - x2
39. g1x2 = - 3x2 - 5
40. h1x2 = 3x3 + 5
3
41. F 1x2 = 2
x
42. G1x2 = 1x
43. f 1x2 = x + 0 x 0
3
44. f 1x2 = 2
2x2 + 1
45. g1x2 =
1
x2
46. h1x2 =
x
x - 1
47. h1x2 =
2
- x3
3x2 - 9
48. F 1x2 =
2x
0x0
In Problems 49–56, for each graph of a function y = f(x), find the absolute maximum and the absolute minimum, if they exist. Identify
any local maximum values or local minimum values.
49.
50.
y
(1, 4)
4
(4, 4)
4
(3, 3)
2
51.
y
(5, 1)
53.
(3, 4)
4
(2, 4)
4
2
(1, 1)
x
5
1
54.
y
4
2
(0, 1)
(1, 1)
(5, 0)
3
5
x
1
55.
y
(2, 4)
4
(1, 3)
(4, 3)
(0, 2)
2
3
y
(0, 3)
(2, 2)
1
52.
y
3
5
x
1
56.
y
(1, 3)
(2, 3)
2
(2, 0)
2
(4, 1)
2
(3, 2)
(0, 2)
1
(0, 2)
(3, 1)
x
3
1 (2, 0) 3
(0, 0)
1
x
y
(3, 2)
2
3
x
3
1
3
x
x
In Problems 57–64, use a graphing utility to graph each function over the indicated interval and approximate any local maximum values
and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places.
57. f 1x2 = x3 - 3x + 2
59. f 1x2 = x5 - x3
1 - 2, 22
58. f 1x2 = x3 - 3x2 + 5
60. f 1x2 = x4 - x2
1 - 2, 22
61. f 1x2 = - 0.2x3 - 0.6x2 + 4x - 6
63. f 1x2 = 0.25x4 + 0.3x3 - 0.9x2 + 3
1 - 6, 42
1 - 3, 22
1 - 1, 32
1 - 2, 22
62. f 1x2 = - 0.4x3 + 0.6x2 + 3x - 2
1 - 4, 52
64. f 1x2 = - 0.4x4 - 0.5x3 + 0.8x2 - 2
1 - 3, 22
234
CHAPTER 3 Functions and Their Graphs
65. Find the average rate of change of f 1x2 = - 2x2 + 4:
(a) From 0 to 2
(b) From 1 to 3
(c) From 1 to 4
66. Find the average rate of change of f 1x2 = - x3 + 1:
(a) From 0 to 2
(b) From 1 to 3
(c) From - 1 to 1
70. f 1x2 = - 4x + 1
(a) Find the average rate of change from 2 to 5.
(b) Find an equation of the secant line containing 12, f 122 2
and 15, f 152 2.
71. g1x2 = x2 - 2
(a) Find the average rate of change from - 2 to 1.
(b) Find an equation of the secant line containing
1 - 2, g1 - 22 2 and 11, g112 2.
67. Find the average rate of change of g1x2 = x3 - 2x + 1:
(a) From - 3 to - 2
(b) From - 1 to 1
(c) From 1 to 3
72. g1x2 = x2 + 1
(a) Find the average rate of change from - 1 to 2.
(b) Find an equation of the secant line containing
1 - 1, g1 - 12 2 and 12, g122 2.
68. Find the average rate of change of h1x2 = x2 - 2x + 3:
(a) From - 1 to 1
(b) From 0 to 2
(c) From 2 to 5
73. h1x2 = x2 - 2x
(a) Find the average rate of change from 2 to 4.
(b) Find an equation of the secant line containing 12, h122 2
and 14, h142 2.
69. f 1x2 = 5x - 2
(a) Find the average rate of change from 1 to 3.
(b) Find an equation of the secant line containing 11, f 112 2
and 13, f 132 2.
74. h1x2 = - 2x2 + x
(a) Find the average rate of change from 0 to 3.
(b) Find an equation of the secant line containing 10, h102 2
and 13, h132 2.
Mixed Practice
75. g(x) = x3 - 27x
(a) Determine whether g is even, odd, or neither.
(b) There is a local minimum value of - 54 at 3.
Determine the local maximum value.
76. f(x) = - x3 + 12x
(a) Determine whether f is even, odd, or neither.
(b) There is a local maximum value of 16 at 2. Determine
the local minimum value.
77. F(x) = - x4 + 8x2 + 8
(a) Determine whether F is even, odd, or neither.
(b) There is a local maximum value of 24 at x = 2. Determine
a second local maximum value.
(c) Suppose the area under the graph of F between
x = 0 and x = 3 that is bounded from below by the
x-axis is 47.4 square units. Using the result from part (a),
determine the area under the graph of F between
x = - 3 and x = 0 that is bounded from below by the
x-axis.
78. G(x) = - x4 + 32x2 + 144
(a) Determine whether G is even, odd, or neither.
(b) There is a local maximum value of 400 at x = 4. Determine
a second local maximum value.
(c) Suppose the area under the graph of G between x = 0
and x = 6 that is bounded from below by the x-axis
is 1612.8 square units. Using the result from part (a),
determine the area under the graph of G between
x = - 6 and x = 0 that is bounded from below by the
x-axis.
Applications and Extensions
79. Minimum Average Cost The average cost per hour in
dollars, C, of producing x riding lawn mowers can be
modeled by the function
C 1x2 = 0.3x2 + 21x - 251 +
2500
x
(a) Use a graphing utility to graph C = C 1x2 .
(b) Determine the number of riding lawn mowers to
produce in order to minimize average cost.
(c) What is the minimum average cost?
80. Medicine Concentration The concentration C of a medication
in the bloodstream t hours after being administered is
modeled by the function
C 1t2 = - 0.002x4 + 0.039t 3 - 0.285t 2 + 0.766t + 0.085
(a) After how many hours will the concentration be
highest?
(b) A woman nursing a child must wait until the concentration
is below 0.5 before she can feed him. After taking the
medication, how long must she wait before feeding her
child?
81. Data Plan Cost The monthly cost C, in dollars, for wireless
data plans with x gigabytes of data included is shown in the
table below. Since each input value for x corresponds to
exactly one output value for C, the plan cost is a function
of the number of data gigabytes. Thus C(x)
represents the monthly cost for a wireless data plan with
x gigabytes included.
GB
Cost ($)
GB
Cost ($)
4
70
20
150
6
80
30
225
10
100
40
300
15
130
50
375
(a) Plot the points (4, 70), (6, 80), (10, 100), and so on in a
Cartesian plane.
(b) Draw a line segment from the point (10, 100) to
(30, 225). What does the slope of this line segment
represent?
SECTION 3.3 Properties of Functions
(c) Find the average rate of change of the monthly cost
from 4 to 10 gigabytes.
(d) Find the average rate of change of the monthly cost
from 10 to 30 gigabytes.
(e) Find the average rate of change of the monthly cost
from 30 to 50 gigabytes.
(f) What is happening to the average rate of change as the
gigabytes of data increase?
82. National Debt The size of the total debt owed by the United
States federal government continues to grow. In fact, according
to the Department of the Treasury, the debt per person
living in the United States is approximately $53,000 (or over
$140,000 per U.S. household). The following data represent
the U.S. debt for the years 2001–2013. Since the debt D
depends on the year y, and each input corresponds to exactly
one output, the debt is a function of the year. So D(y)
represents the debt for each year y.
Year
Debt (billions
of dollars)
Year
Debt (billions
of dollars)
2001
5807
2008
10,025
2002
6228
2009
11,910
2003
6783
2010
13,562
2004
7379
2011
14,790
2005
7933
2012
16,066
2006
8507
2013
16,738
2007
9008
(a) Find the average rate of change of the population from
0 to 2.5 hours.
(b) Find the average rate of change of the population from
4.5 to 6 hours.
(c) What is happening to the average rate of change as time
passes?
84. e-Filing Tax Returns The Internal Revenue Service
Restructuring and Reform Act (RRA) was signed into
law by President Bill Clinton in 1998. A major objective of
the RRA was to promote electronic filing of tax returns.
The data in the table that follows show the percentage of
individual income tax returns filed electronically for filing
years 2004–2012. Since the percentage P of returns filed
electronically depends on the filing year y, and each input
corresponds to exactly one output, the percentage of
returns filed electronically is a function of the filing year; so
P 1y2 represents the percentage of returns filed electronically
for filing year y.
(a) Find the average rate of change of the percentage of
e-filed returns from 2004 to 2006.
(b) Find the average rate of change of the percentage of
e-filed returns from 2007 to 2009.
(c) Find the average rate of change of the percentage of
e-filed returns from 2010 to 2012.
(d) What is happening to the average rate of change as time
passes?
Source: www.treasurydirect.gov
(a) Plot the points (2001, 5807), (2002, 6228), and so on in a
Cartesian plane.
(b) Draw a line segment from the point (2001, 5807) to
(2006, 8507). What does the slope of this line segment
represent?
(c) Find the average rate of change of the debt from 2002 to
2004.
(d) Find the average rate of change of the debt from 2006 to
2008.
(e) Find the average rate of change of the debt from 2010 to
2012.
(f) What appears to be happening to the average rate of
change as time passes?
83. E. coli Growth A strain of E. coli Beu 397-recA441 is placed
into a nutrient broth at 30° Celsius and allowed to grow. The
data shown in the table are collected. The population is
measured in grams and the time in hours. Since population P
depends on time t, and each input corresponds to exactly
one output, we can say that population is a function of time.
Thus P 1t2 represents the population at time t.
Time
(hours), t
Population
(grams), P
0
0.09
2.5
0.18
3.5
0.26
4.5
0.35
6
0.50
235
Year
Percentage of
returns e-filed
2004
46.5
2005
51.1
2006
53.8
2007
57.1
2008
58.5
2009
67.2
2010
69.8
2011
77.2
2012
82.7
Source: Internal Revenue Service
85. For the function f 1x2 = x2, compute the average rate of
change:
(a) From 0 to 1
(b) From 0 to 0.5
(c) From 0 to 0.1
(d) From 0 to 0.01
(e) From 0 to 0.001
(f) Use a graphing utility to graph each of the secant lines
along with f .
(g) What do you think is happening to the secant lines?
(h) What is happening to the slopes of the secant lines? Is
there some number that they are getting closer to? What
is that number?
236
CHAPTER 3 Functions and Their Graphs
86. For the function f 1x2 = x2, compute the average rate of
change:
(a) From 1 to 2
(b) From 1 to 1.5
(c) From 1 to 1.1
(d) From 1 to 1.01
(e) From 1 to 1.001
(f) Use a graphing utility to graph each of the secant lines
along with f .
(g) What do you think is happening to the secant lines?
(h) What is happening to the slopes of the secant lines? Is
there some number that they are getting closer to? What
is that number?
Problems 87–94 require the following discussion of a secant line. The slope of the secant line containing the two points 1x, f 1x2 2 and
1x + h, f 1x + h2 2 on the graph of a function y = f 1x2 may be given as
msec =
f 1x + h2 - f 1x2
1x + h2 - x
=
f 1x + h2 - f 1x2
h
,
h ≠ 0
In calculus, this expression is called the difference quotient of f.
(a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer.
(b) Find msec for h = 0.5, 0.1, and 0.01 at x = 1. What value does msec approach as h approaches 0?
(c) Find an equation for the secant line at x = 1 with h = 0.01.
(d) Use a graphing utility to graph f and the secant line found in part (c) in the same viewing window.
87. f 1x2 = 2x + 5
88. f 1x2 = - 3x + 2
89. f 1x2 = x2 + 2x
90. f 1x2 = 2x2 + x
91. f 1x2 = 2x2 - 3x + 1
92. f 1x2 = - x2 + 3x - 2
93 f 1x2 =
94. f 1x2 =
1
x
1
x2
Explaining Concepts: Discussion and Writing
95. Draw the graph of a function that has the following properties:
domain: all real numbers; range: all real numbers; intercepts:
10, - 32 and 13, 02; a local maximum value of - 2 is at - 1; a
local minimum value of - 6 is at 2. Compare your graph with
those of others. Comment on any differences.
96. Redo Problem 95 with the following additional information:
increasing on 1 - q , - 12, 12, q 2; decreasing on 1 - 1, 22.
Again compare your graph with others and comment on any
differences.
97. How many x-intercepts can a function defined on an interval
have if it is increasing on that interval? Explain.
99. Can a function be both even and odd? Explain.
100. Using a graphing utility, graph y = 5 on the interval
1 - 3, 32. Use MAXIMUM to find the local maximum
values on 1 - 3, 32. Comment on the result provided by the
calculator.
101. A function f has a positive average rate of change on the
interval 3 2, 5 4 . Is f increasing on 3 2, 5 4 ? Explain.
102. Show that a constant function f(x) = b has an average
rate of change of 0. Compute the average rate of change of
y = 24 - x2 on the interval 3 - 2, 24. Explain how this can
happen.
98. Suppose that a friend of yours does not understand the idea of
increasing and decreasing functions. Provide an explanation,
complete with graphs, that clarifies the idea.
Retain Your Knowledge
Problems 103–106 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
103. Write each number in scientific notation.
(a) 0.00000701
(b) 2, 305, 000, 000
104. Simplify: 2540
105. Solve: 14 6 5 - 3x … 29
106. The shelf life of a perishable commodity varies inversely
with the storage temperature. If the shelf life at 10°C is
33 days, what is the shelf life at 40°C?
‘Are You Prepared?’ Answers
1. 2 6 x 6 5
2. 1
3. symmetric with respect to the y-axis
4. y + 2 = 51x - 32
5. 1 - 3, 02, 13, 02, 10, - 92
SECTION 3.4 Library of Functions; Piecewise-defined Functions
237
3.4 Library of Functions; Piecewise-defined Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
r Intercepts (Section 2.2, pp. 159–160)
r Graphs of Key Equations (Section 2.2:
Example 3, p. 159; Example 10, p. 163;
Example 11, p. 163; Example 12, p. 164)
Now Work the ‘Are You Prepared?’ problems on page 244.
OBJECTIVES 1 Graph the Functions Listed in the Library of Functions (p. 237)
2 Graph Piecewise-defined Functions (p. 242)
1 Graph the Functions Listed in the Library of Functions
First we introduce a few more functions, beginning with the square root function.
On page 163, we graphed the equation y = 1x. Figure 28 shows a graph of the
function f1x2 = 1x. Based on the graph, we have the following properties:
Properties of f (x) = !x
y
6
(1, 1)
(4, 2)
(9, 3)
(0, 0)
–2
5
10 x
Figure 28 Square root function
EXAMPL E 1
1. The domain and the range are the set of nonnegative real numbers.
2. The x-intercept of the graph of f1x2 = 1x is 0. The y-intercept of the
graph of f1x2 = 1x is also 0.
3. The function is neither even nor odd.
4. The function is increasing on the interval 10, q 2.
5. The function has an absolute minimum of 0 at x = 0.
Graphing the Cube Root Function
3
(a) Determine whether f1x2 = 2
x is even, odd, or neither. State whether the
graph of f is symmetric with respect to the y-axis or symmetric with respect to
the origin.
3
(b) Determine the intercepts, if any, of the graph of f1x2 = 2
x.
3
(c) Graph f1x2 = 2
x.
Solution
(a) Because
3
3
f 1 - x2 = 2
-x = - 2
x = - f1x2
the function is odd. The graph of f is symmetric with respect to the origin.
3
(b) The y-intercept is f102 = 2
0 = 0. The x-intercept is found by solving the
equation f 1x2 = 0.
f1x2 = 0
3
2
x = 0
x = 0
3
f (x) = 2
x
Cube both sides of the equation.
The x-intercept is also 0.
(c) Use the function to form Table 4 (on page 238) and obtain some points on the
graph. Because of the symmetry with respect to the origin, we find only points
3
1x, y2 for which x Ú 0. Figure 29 shows the graph of f1x2 = 2x.
238
CHAPTER 3 Functions and Their Graphs
Table 4
x
y = f (x) = !x
0
0
(0, 0)
1
8
1
2
1 1
a , b
8 2
1
1
(1, 1)
3
2
3
(
1–8 , 1–2
2
(1, 1)
)
(2, 2 )
( 1–8 , 1–2)
3
3
1 2, 2
22
3
2
2 ≈ 1.26
8
y
3
(x, y)
3
x
(0, 0)
3
(1, 1)
(2, 2 )
(8, 2)
3
Figure 29 Cube Root Function
r
From the results of Example 1 and Figure 29, we have the following properties
of the cube root function.
Properties of f(x) = !x
3
1. The domain and the range are the set of all real numbers.
3
2. The x-intercept of the graph of f1x2 = 2
x is 0. The y-intercept of the
3
graph of f1x2 = 2
x is also 0.
3. The function is odd. The graph is symmetric with respect to the origin.
4. The function is increasing on the interval 1 - q , q 2.
5. The function does not have any local minima or any local maxima.
EX AMPLE 2
Solution
Graphing the Absolute Value Function
(a) Determine whether f 1x2 = 0 x 0 is even, odd, or neither. State whether the
graph of f is symmetric with respect to the y-axis, symmetric with respect to the
origin, or neither.
(b) Determine the intercepts, if any, of the graph of f1x2 = 0 x 0 .
(c) Graph f1x2 = 0 x 0 .
(a) Because
f1 - x2 = 0 - x 0
= 0 x 0 = f1x2
the function is even. The graph of f is symmetric with respect to the y-axis.
(b) The y-intercept is f102 = 0 0 0 = 0. The x-intercept is found by solving the
equation f 1x2 = 0, or 0 x 0 = 0. The x-intercept is 0.
(c) Use the function to form Table 5 and obtain some points on the graph. Because
of the symmetry with respect to the y-axis, we only need to find points 1x, y2 for
which x Ú 0. Figure 30 shows the graph of f1x2 = 0 x 0 .
Table 5
x
y = f (x) = ∣ x ∣
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
2
(2, 2)
3
3
(3, 3)
y
3
(3, 3)
(2, 2)
(1, 1)
3 2 1
(3, 3)
(2, 2)
2
1
1
(1, 1)
1 2
(0, 0)
3
Figure 30 Absolute Value Function
x
r
From the results of Example 2 and Figure 30, we have the following properties
of the absolute value function.
SECTION 3.4 Library of Functions; Piecewise-defined Functions
239
Properties of f(x) = ∣ x ∣
1. The domain is the set of all real numbers. The range of f is 5 y y Ú 06 .
2. The x-intercept of the graph of f 1x2 = 0 x 0 is 0. The y-intercept of the
graph of f 1x2 = 0 x 0 is also 0.
3. The function is even. The graph is symmetric with respect to the y-axis.
4. The function is decreasing on the interval 1 - q , 02. It is increasing on the
interval 10, q 2.
5. The function has an absolute minimum of 0 at x = 0.
Seeing the Concept
Graph y = 0 x 0 on a square screen and compare what you see with Figure 30. Note that some graphing
calculators use abs(x) for absolute value.
Below is a list of the key functions that we have discussed. In going through this list,
pay special attention to the properties of each function, particularly to the shape of
each graph. Knowing these graphs, along with key points on each graph, will lay the
foundation for further graphing techniques.
y
Constant Function
f(x) = b
(0,b)
f1x2 = b
b is a real number
x
Figure 31 Constant Function
See Figure 31.
The domain of a constant function is the set of all real numbers; its range is the
set consisting of a single number b. Its graph is a horizontal line whose y-intercept
is b. The constant function is an even function.
f (x ) = x
y
3
Identity Function
f1x2 = x
(1, 1)
(0, 0)
–3
(– 1, –1)
3 x
See Figure 32.
The domain and the range of the identity function are the set of all real
numbers. Its graph is a line whose slope is 1 and whose y-intercept is 0. The line
consists of all points for which the x-coordinate equals the y-coordinate. The identity
function is an odd function that is increasing over its domain. Note that the graph
bisects quadrants I and III.
Figure 32 Identity Function
y
(–2, 4)
4
(–1, 1)
–4
f (x ) = x 2
Square Function
(2, 4)
f1x2 = x2
(1, 1)
(0, 0)
Figure 33 Square Function
4 x
See Figure 33.
The domain of the square function is the set of all real numbers; its range is
the set of nonnegative real numbers. The graph of this function is a parabola whose
intercept is at 10, 02. The square function is an even function that is decreasing on
the interval 1 - q , 02 and increasing on the interval 10, q 2.
240
CHAPTER 3 Functions and Their Graphs
y
Cube Function
4
f(x) = x 3
f1x2 = x3
(1, 1)
4
(1, 1)
(0, 0)
x
4
See Figure 34.
The domain and the range of the cube function are the set of all real numbers.
The intercept of the graph is at 10, 02. The cube function is odd and is increasing
on the interval 1 - q , q 2.
4
Figure 34 Cube Function
y
Square Root Function
f(x) = x
2
(1, 1)
1
f1x2 = 2x
(4, 2)
5 x
(0, 0)
Figure 35 Square Root Function
y
3
3
(
1–8 , 1–2
(1, 1)
)
(2, 2 )
See Figure 35.
The domain and the range of the square root function are the set of nonnegative
real numbers. The intercept of the graph is at 10, 02. The square root function is
neither even nor odd and is increasing on the interval 10, q 2.
Cube Root Function
3
f 1x2 = 2
x
( 1–8 , 1–2)
3
3 x
(0, 0)
3
(1, 1)
(2, 2 )
3
Figure 36 Cube Root Function
y
2
Reciprocal Function
(1–2 , 2)
f(x ) =
(2, 1–2 )
See Figure 36.
The domain and the range of the cube root function are the set of all real
numbers. The intercept of the graph is at 10, 02. The cube root function is an odd
function that is increasing on the interval 1 - q , q 2.
f1x2 =
1
––
x
1
x
(1, 1)
2
2 x
(1, 1)
2
Figure 37 Reciprocal Function
1
Refer to Example 12, page 164, for a discussion of the equation y = . See
x
Figure 37.
The domain and the range of the reciprocal function are the set of all nonzero
real numbers. The graph has no intercepts. The reciprocal function is decreasing on
the intervals 1 - q , 02 and 10, q 2 and is an odd function.
Absolute Value Function
y
f(x) = ⏐x ⏐
f1x2 = 0 x 0
3
(2, 2)
(2, 2)
(1, 1)
3
(0, 0)
(1, 1)
3 x
Figure 38 Absolute Value Function
See Figure 38.
The domain of the absolute value function is the set of all real numbers; its
range is the set of nonnegative real numbers. The intercept of the graph is at 10, 02.
If x Ú 0, then f 1x2 = x, and the graph of f is part of the line y = x; if x 6 0,
then f1x2 = - x, and the graph of f is part of the line y = - x. The absolute value
function is an even function; it is decreasing on the interval 1 - q , 02 and increasing
on the interval 10, q 2.
SECTION 3.4 Library of Functions; Piecewise-defined Functions
241
The notation int1x2 stands for the largest integer less than or equal to x. For
example,
1
3
int112 = 1, int12.52 = 2, inta b = 0, inta - b = - 1, int1p2 = 3
2
4
This type of correspondence occurs frequently enough in mathematics that we give
it a name.
Table 6
y = f (x)
x
(x, y)
= int (x)
DEFINITION Greatest Integer Function
-1
-1
( - 1, - 1)
1
2
-1
1
a - , - 1b
2
1
4
-1
1
a - , - 1b
4
0
0
(0, 0)
1
4
0
1
a , 0b
4
1
2
0
1
a , 0b
2
3
4
0
3
a , 0b
4
-
y
4
2
2
2
4
x
3
Figure 39 Greatest Integer Function
f1x2 = int1x2* = greatest integer less than or equal to x
We obtain the graph of f1x2 = int1x2 by plotting several points. See Table 6. For
values of x, - 1 … x 6 0, the value of f 1x2 = int 1x2 is - 1; for values of x,
0 … x 6 1, the value of f is 0. See Figure 39 for the graph.
The domain of the greatest integer function is the set of all real numbers; its range
is the set of integers. The y-intercept of the graph is 0. The x-intercepts lie in the interval
3 0, 12. The greatest integer function is neither even nor odd. It is constant on every
interval of the form 3 k, k + 12, for k an integer. In Figure 39, a solid dot is used to
indicate, for example, that at x = 1 the value of f is f112 = 1; an open circle is used to
illustrate that the function does not assume the value of 0 at x = 1.
Although a precise definition requires the idea of a limit (discussed in calculus),
in a rough sense, a function is said to be continuous if its graph has no gaps or holes
and can be drawn without lifting a pencil from the paper on which the graph is
drawn. We contrast this with a discontinuous function. A function is discontinuous
if its graph has gaps or holes and so cannot be drawn without lifting a pencil from
the paper.
From the graph of the greatest integer function, we can see why it is also called a
step function. At x = 0, x = {1, x = {2, and so on, this function is discontinuous
because, at integer values, the graph suddenly “steps” from one value to another
without taking on any of the intermediate values. For example, to the immediate
left of x = 3, the y-coordinates of the points on the graph are 2, and at x = 3 and
to the immediate right of x = 3, the y-coordinates of the points on the graph are 3.
Consequently, the graph has gaps in it.
COMMENT When graphing a function using a graphing utility, typically you can choose
either connected mode, in which points plotted on the screen are connected, making the graph appear
without any breaks, or dot mode, in which only the points plotted appear. When graphing the greatest
integer function with a graphing utility, it may be necessary to be in dot mode. This is to prevent the
utility from “connecting the dots” when f1x2 changes from one integer value to the next. However, some
utilities will display the gaps even when in “connected” mode. See Figure 40.
■
6
6
6
22
6
22
Figure 40
f 1x2 = int 1x2
D TI-83 Plus, connected mode
22
6
22
E TI-83 Plus, dot mode
*Some texts use the notation f 1x2 = 3 x 4 instead of int 1x2.
−2
6
−2
(c) TI-84 Plus C
242
CHAPTER 3 Functions and Their Graphs
The functions discussed so far are basic. Whenever you encounter one of them,
you should see a mental picture of its graph. For example, if you encounter the
function f1x2 = x2, you should see in your mind’s eye a picture like Figure 33.
Now Work
PROBLEMS
11
THROUGH
18
2 Graph Piecewise-defined Functions
Sometimes a function is defined using different equations on different parts of its
domain. For example, the absolute value function f1x2 = 0 x 0 is actually defined by
two equations: f 1x2 = x if x Ú 0 and f1x2 = - x if x 6 0. For convenience, these
equations are generally combined into one expression as
f1x2 = 0 x 0 = e
x if x Ú 0
- x if x 6 0
When a function is defined by different equations on different parts of its domain,
it is called a piecewise-defined function.
EX AMPLE 3
Analyzing a Piecewise-defined Function
The function f is defined as
- 2x + 1 if - 3 … x 6 1
f1x2 = c 2
if x = 1
x2
if x 7 1
(a) Find f1 - 22 , f112, and f122.
(c) Locate any intercepts.
(e) Use the graph to find the range of f.
Solution
(b) Determine the domain of f.
(d) Graph f .
(f) Is f continuous on its domain?
(a) To find f1 - 22, observe that when x = - 2, the equation for f is given by
f1x2 = - 2x + 1. So
f1 - 22 = - 2( - 2) + 1 = 5
When x = 1, the equation for f is f 1x2 = 2. So,
f112 = 2
When x = 2, the equation for f is f 1x2 = x2. So
f122 = 22 = 4
(b) To find the domain of f, look at its definition. Since f is defined for all x greater
than or equal to - 3, the domain of f is 5 x x Ú - 36 , or the interval 3 - 3, q 2 .
(c) The y-intercept of the graph of the function is f 102 . Because the equation for
f when x = 0 is f1x2 = - 2x + 1, the y-intercept is f102 = - 2102 + 1 = 1.
The x-intercepts of the graph of a function f are the real solutions to the
equation f1x2 = 0. To find the x-intercepts of f, solve f1x2 = 0 for each “piece”
of the function, and then determine what values of x, if any, satisfy the condition
that defines the piece.
f1x2 = 0
- 2x + 1 = 0
-3 … x 6 1
- 2x = - 1
1
x =
2
f1x2 = 0
2 = 0
x = 1
No solution
f1x2 = 0
x2 = 0
x = 0
x 7 1
SECTION 3.4 Library of Functions; Piecewise-defined Functions
y
8
4
(1,2)
(2,4)
(0,1)
4
( 1–2 , 0)
(1,
x
1)
243
1
The first potential x-intercept, x = , satisfies the condition - 3 … x 6 1, so
2
1
x = is an x-intercept. The second potential x-intercept, x = 0, does not satisfy
2
1
the condition x 7 1, so x = 0 is not an x-intercept. The only x-intercept is . The
2
1
intercepts are (0, 1) and a , 0b .
2
(d) To graph f , graph each “piece.” First graph the line y = - 2x + 1 and keep
only the part for which - 3 … x 6 1. Then plot the point 11, 22 because, when
x = 1, f1x2 = 2. Finally, graph the parabola y = x2 and keep only the part for
which x 7 1. See Figure 41.
(e) From the graph, we conclude that the range of f is 5 y y 7 - 16 , or the interval
1 - 1, q 2.
(f) The function f is not continuous because there is a “jump” in the graph at
x = 1.
r
Figure 41
Now Work
EXAMPL E 4
PROBLEM
31
Cost of Electricity
In the spring of 2014, Duke Energy Progress supplied electricity to residences in
South Carolina for a monthly customer charge of $6.50 plus 9.971¢ per kilowatt-hour
(kWh) for the first 800 kWh supplied in the month and 8.971¢ per kWh for all usage
over 800 kWh in the month.
(a) What is the charge for using 300 kWh in a month?
(b) What is the charge for using 1500 kWh in a month?
(c) If C is the monthly charge for x kWh, develop a model relating the monthly
charge and kilowatt-hours used. That is, express C as a function of x.
Source: Duke Energy Progress, 2014
Solution
(a) For 300 kWh, the charge is $6.50 plus (9.971c = $0.09971) per kWh. That is,
Charge = $6.50 + $0.09971(300) = $36.41
(b) For 1500 kWh, the charge is $6.50 plus 9.971c per kWh for the first 800 kWh plus
8.971c per kWh for the 700 in excess of 800. That is,
Charge = $6.50 + $0.09971(800) + $0.08971(700) = $ 149.07
(c) Let x represent the number of kilowatt-hours used. If 0 … x … 800, then the
monthly charge C (in dollars) can be found by multiplying x times $0.09971and
adding the monthly customer charge of $6.50. So if 0 … x … 800, then
C(x) = 0.09971x + 6.50
For x 7 800, the charge is 0.09971(800) + 6.50 + 0.08971(x - 800),
since (x - 800) equals the usage in excess of 800 kWh, which costs $0.08971 per
kWh. That is, if x 7 800, then
C
C(x) = 0.09971(800) + 6.50 + 0.08971(x - 800)
= 79.768 + 6.50 + 0.08971x - 71.768
= 0.08971x + 14.50
Charge (dollars)
180
(1500, 149.07)
120
60
The rule for computing C follows two equations:
(800, 86.27)
(0, 6.50)
C 1x2 = e
(300, 36.41)
400
Figure 42
800
1200
Usage (kWh)
x
0.09971x + 6.50 if 0 … x … 800
0.08971x + 14.50 if x 7 800
The Model
See Figure 42 for the graph. Note that the two “pieces” are linear, but they have
different slopes (rates), and meet at the point (800, 86.27).
r
244
CHAPTER 3 Functions and Their Graphs
3.4 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Sketch the graph of y = 1x. (p. 163)
2. Sketch the graph of y =
3. List the intercepts of the equation y = x3 - 8. (pp. 159–160)
1
. (p. 164)
x
Concepts and Vocabulary
4. The function f 1x2 = x2 is decreasing on the interval
.
5. When functions are defined by more than one equation,
functions.
they are called
6. True or False The cube function is odd and is increasing on
the interval 1 - q , q 2.
9. Which of the following functions has a graph that is
symmetric about the y-axis?
1
(a) y = 2x (b) y = x (c) y = x3 (d) y =
x
10. Consider the following function.
3x - 2
f(x) = c x2 + 5
3
7. True or False The cube root function is odd and is
decreasing on the interval 1 - q , q 2.
8. True or False The domain and the range of the reciprocal
function are the set of all real numbers.
x 6 2
2 … x 6 10
x Ú 10
if
if
if
Which “piece(s)” should be used to find the y-intercept?
(a) 3x - 2 (b) x2 + 5 (c) 3 (d) all three
Skill Building
In Problems 11–18, match each graph to its function.
A. Constant function
E. Square root function
B. Identity function
F. Reciprocal function
C. Square function
G. Absolute value function
D. Cube function
H. Cube root function
11.
12.
13.
14.
15.
16.
17.
18.
In Problems 19–26, sketch the graph of each function. Be sure to label three points on the graph.
19. f 1x2 = x
20. f 1x2 = x2
21. f 1x2 = x3
22. f 1x2 = 1x
23. f 1x2 =
24. f 1x2 = 0 x 0
3
25. f 1x2 = 2x
26. f 1x2 = 3
1
x
x2
27. If f 1x2 = c 2
2x + 1
find: (a) f 1 - 22
29. If f 1x2 = e
2x - 4
x3 - 2
find: (a) f 102
if x 6 0
if x = 0
if x 7 0
(b) f 102
- 3x
28. If f 1x2 = c 0
2x2 + 1
(c) f 122
if - 1 … x … 2
if 2 6 x … 3
(b) f 112
find: (a) f 1 - 22
30. If f 1x2 = e
(c) f 122
(d) f 132
x3
3x + 2
find: (a) f 1 - 12
if x 6 - 1
if x = - 1
if x 7 - 1
(b) f 1 - 12
(c) f 102
if - 2 … x 6 1
if 1 … x … 4
(b) f 102
(c) f 112
(d) f(3)
In Problems 31–42:
(a) Find the domain of each function.
(d) Based on the graph, find the range.
31. f 1x2 = b
2x
1
if x ≠ 0
if x = 0
(b) Locate any intercepts.
(e) Is f continuous on its domain?
32. f 1x2 = b
3x
4
if x ≠ 0
if x = 0
(c) Graph each function.
33. f 1x2 = b
- 2x + 3
3x - 2
if x 6 1
if x Ú 1
SECTION 3.4 Library of Functions; Piecewise-defined Functions
34. f 1x2 = b
x + 3
- 2x - 3
37. f 1x2 = b
1 + x
x2
40. f 1x2 = b
x + 3
35. f 1x2 = c 5
-x + 2
if x 6 - 2
if x Ú - 2
1
if x 6 0
if x Ú 0
38. f 1x2 = c x
3
2
x
2 - x
if - 3 … x 6 1
2x
if x 7 1
41. f 1x2 = 2 int 1x2
if - 2 … x 6 1
if x = 1
if x 7 1
if x 6 0
if - 3 … x 6 0
if x = 0
if x 7 0
2x + 5
36. f 1x2 = c - 3
- 5x
39. f 1x2 = b
if x Ú 0
0x0
x
245
if - 2 … x 6 0
3
if x 7 0
42. f 1x2 = int 12x2
In Problems 43–46, the graph of a piecewise-defined function is given. Write a definition for each function.
y
43.
y
2
44.
2
(2, 2)
(1, 1)
(1, 1)
2
(0, 0)
2 x
2
y
(0, 2)
46.
2
(2, 1)
(2, 1)
(1, 1)
y
45.
(0, 0)
2 x
2
(1, 0)
(0, 0)
(2, 0) x
2
(1, 1)
2 x
(1, 1)
47. If f 1x2 = int 12x2, find
(a) f 11.22
(b) f 11.62
(c) f 1 - 1.82
x
48. If f 1x2 = int a b, find
2
(a) f 11.22
(b) f 11.62
(c) f 1 - 1.82
Applications and Extensions
49. Tablet Service Sprint offers a monthly tablet plan for $34.99.
It includes 3 gigabytes of data and charges $15 per gigabyte
for additional gigabytes. The following function is used to
compute the monthly cost for a subscriber.
34.99
C(x) = e
15x - 10.01
if
if
0 … x … 3
x 7 3
Compute the monthly cost for each of the following
gigabytes of use.
(a) 2
(b) 5
(c) 13
Source: Sprint, April 2014
50. Parking at O’Hare International Airport The short-term
(no more than 24 hours) parking fee F (in dollars) for parking
x hours on a weekday at O’Hare International Airport’s
main parking garage can be modeled by the function
3
F 1x2 = c 5 int 1x + 12 + 1
50
if 0 6 x … 3
if 3 6 x 6 9
if 9 … x … 24
Determine the fee for parking in the short-term parking
garage for
(a) 2 hours
(b) 7 hours
(c) 15 hours
(d) 8 hours and 24 minutes
Source: O’Hare International Airport
51. Cost of Natural Gas In March 2014, Laclede Gas had
the rate schedule (on, right) for natural gas usage in
single-family residences.
(a) What is the charge for using 20 therms in a month?
(b) What is the charge for using 150 therms in a month?
(c) Develop a function that models the monthly charge C
for x therms of gas.
(d) Graph the function found in part (c).
Monthly service charge
Delivery charge
First 30 therms
Over 30 therms
Natural gas cost
First 30 therms
Over 30 therms
$19.50
$0.91686/therm
$0
$0.3313/therm
$0.5757/therm
Source: Laclede Gas
52. Cost of Natural Gas In April 2014, Nicor Gas had the
following rate schedule for natural gas usage in small
businesses.
Monthly customer charge
Distribution charge
1st 150 therms
Next 4850 therms
Over 5000 therms
Gas supply charge
$72.60
$0.1201/therm
$0.0549/therm
$0.0482/therm
$0.68/therm
(a) What is the charge for using 1000 therms in a month?
(b) What is the charge for using 6000 therms in a month?
(c) Develop a function that models the monthly charge C
for x therms of gas.
(d) Graph the function found in part (c).
Source: Nicor Gas, 2014
246
CHAPTER 3 Functions and Their Graphs
53. Federal Income Tax Two 2014 Tax Rate Schedules are given in the accompanying table. If x equals taxable income and y equals
the tax due, construct a function y = f 1x2 for Schedule X.
2014 Tax Rate Schedules
Schedule X—Single
If Taxable
Income is
Over
Schedule Y-1—Married Filing Jointly or Qualified Widow(er)
But Not
Over
The Tax is
This
Amount
$0
$9,075
$0
+
10%
$0
9,075
36,900
907.50
+
15%
9,075
Plus This
%
Of the
Excess
Over
If Taxable
Income is
Over
But Not
Over
The Tax is
This
Amount
$0
$18,150
$0
+
10%
$0
18,150
73,800
1,815
+
15%
18,150
Plus This
%
Of the
Excess
Over
36,900
89,350
5,081.25
+
25%
36,900
73,800
148,850
10,162.50
+
25%
73,800
89,350
186,350
18,193.75
+
28%
89,350
148,850
226,850
28,925.00
+
28%
148,850
186,350
405,100
45,353.75
+
33%
186,350
226,850
405,100
50,765.00
+
33%
226,850
405,100
406,750
117,541.25
+
35%
405,100
405,100
457,600
109,587.50
+
35%
405,100
406,750
-
118,188.75
+
39.6%
406,750
457,600
-
127,962.50
+
39.6%
457,600
54. Federal Income Tax Refer to the 2014 tax rate schedules. If
x equals taxable income and y equals the tax due, construct
a function y = f 1x2 for Schedule Y-1.
55. Cost of Transporting Goods
A trucking company
transports goods between Chicago and New York, a distance
of 960 miles. The company’s policy is to charge, for each
pound, $0.50 per mile for the first 100 miles, $0.40 per mile
for the next 300 miles, $0.25 per mile for the next 400 miles,
and no charge for the remaining 160 miles.
(a) Graph the relationship between the cost of transportation
in dollars and mileage over the entire 960-mile route.
(b) Find the cost as a function of mileage for hauls between
100 and 400 miles from Chicago.
(c) Find the cost as a function of mileage for hauls between
400 and 800 miles from Chicago.
56. Car Rental Costs An economy car rented in Florida from
Enterprise® on a weekly basis costs $185 per week. Extra
days cost $37 per day until the day rate exceeds the weekly
rate, in which case the weekly rate applies. Also, any part of
a day used counts as a full day. Find the cost C of renting
an economy car as a function of the number x of days used,
where 7 … x … 14. Graph this function.
57. Mortgage Fees Fannie Mae charges a loan-level price
adjustment (LLPA) on all mortgages, which represents a fee
homebuyers seeking a loan must pay. The rate paid depends
on the credit score of the borrower, the amount borrowed,
and the loan-to-value (LTV) ratio. The LTV ratio is the ratio
of amount borrowed to appraised value of the home. For
example, a homebuyer who wishes to borrow $250,000 with
a credit score of 730 and an LTV ratio of 80% will pay 0.5%
(0.005) of $250,000, or $1250. The table shows the LLPA for
various credit scores and an LTV ratio of 80%.
Credit Score
Loan-Level
Price Adjustment Rate
… 659
3.00%
660–679
2.50%
680–699
1.75%
700–719
1%
720–739
0.5%
Ú 740
0.25%
Source: Fannie Mae.
(a) Construct a function C = C(s), where C is the loan-level
price adjustment (LLPA) and s is the credit score of an
individual who wishes to borrow $300,000 with an 80%
LTV ratio.
(b) What is the LLPA on a $300,000 loan with an 80% LTV
ratio for a borrower whose credit score is 725?
(c) What is the LLPA on a $300,000 loan with an 80% LTV
ratio for a borrower whose credit score is 670?
58. Minimum Payments for Credit Cards Holders of credit
cards issued by banks, department stores, oil companies, and
so on receive bills each month that state minimum amounts
that must be paid by a certain due date. The minimum due
depends on the total amount owed. One such credit card
company uses the following rules: For a bill of less than $10,
the entire amount is due. For a bill of at least $10 but less
than $500, the minimum due is $10. A minimum of $30 is
due on a bill of at least $500 but less than $1000, a minimum
of $50 is due on a bill of at least $1000 but less than $1500,
and a minimum of $70 is due on bills of $1500 or more. Find
the function f that describes the minimum payment due on
a bill of x dollars. Graph f.
59. Wind Chill The wind chill factor represents the air
temperature at a standard wind speed that would produce
the same heat loss as the given temperature and wind speed.
One formula for computing the equivalent temperature is
t
W = d 33 -
110.45 + 101v - v2 133 - t2
22.04
33 - 1.5958133 - t2
0 … v 6 1.79
1.79 … v … 20
v 7 20
where v represents the wind speed (in meters per second)
and t represents the air temperature (°C). Compute the wind
chill for the following:
(a) An air temperature of 10°C and a wind speed of 1 meter
per second 1m/sec2
(b) An air temperature of 10°C and a wind speed of
5 m/sec
(c) An air temperature of 10°C and a wind speed of
15 m/sec
(d) An air temperature of 10°C and a wind speed of 25 m/sec
(e) Explain the physical meaning of the equation
corresponding to 0 … v 6 1.79.
(f) Explain the physical meaning of the equation
corresponding to v 7 20.
SECTION 3.5 Graphing Techniques: Transformations
60. Wind Chill Redo Problem 59(a)–(d) for an air temperature
of - 10°C.
61. First-class Mail In 2014 the U.S. Postal Service charged
$0.98 postage for first-class mail retail flats (such as an 8.5"
by 11" envelope) weighing up to 1 ounce, plus $0.21 for each
247
additional ounce up to 13 ounces. First-class rates do not
apply to flats weighing more than 13 ounces. Develop a
model that relates C, the first-class postage charged, for a
flat weighing x ounces. Graph the function.
Source: United States Postal Service
Explaining Concepts: Discussion and Writing
In Problems 62–69, use a graphing utility.
62. Exploration Graph y = x2. Then on the same screen
graph y = x2 + 2, followed by y = x2 + 4, followed by
y = x2 - 2. What pattern do you observe? Can you predict
the graph of y = x2 - 4? Of y = x2 + 5?
63. Exploration Graph y = x2. Then on the same screen
graph y = 1x - 22 2, followed by y = 1x - 42 2, followed
by y = 1x + 22 2. What pattern do you observe? Can you
predict the graph of y = 1x + 42 2? Of y = 1x - 52 2?
64. Exploration Graph y = 0 x 0 . Then on the same screen graph
1
y = 2 0 x 0 , followed by y = 4 0 x 0 , followed by y = 0 x 0 .
2
What pattern do you observe? Can you predict the graph of
1
y = 0 x 0 ? Of y = 5 0 x 0 ?
4
65. Exploration Graph y = x2. Then on the same screen graph
y = - x2. Now try y = 0 x 0 and y = - 0 x 0 . What do you
conclude?
66. Exploration Graph y = 1x. Then on the same screen graph
y = 1 - x. Now try y = 2x + 1 and y = 21 - x2 + 1. What
do you conclude?
67. Exploration Graph y = x3. Then on the same screen graph
y = 1x - 12 3 + 2. Could you have predicted the result?
68. Exploration Graph y = x2, y = x4, and y = x6 on the same
screen. What do you notice is the same about each graph?
What do you notice is different?
69. Exploration Graph y = x3, y = x5, and y = x7 on the same
screen. What do you notice is the same about each graph?
What do you notice is different?
70. Consider the equation
y = b
1
0
if x is rational
if x is irrational
Is this a function? What is its domain? What is its range?
What is its y-intercept, if any? What are its x-intercepts, if
any? Is it even, odd, or neither? How would you describe its
graph?
71. Define some functions that pass through 10, 02 and
11, 12 and are increasing for x Ú 0. Begin your list with
y = 1x, y = x, and y = x2. Can you propose a general
result about such functions?
Retain Your Knowledge
Problems 72–75 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
72. Simplify: 13 + 2i2 14 - 5i2
73. Find the center and radius of the circle x + y = 6y + 16.
2
2
74. Solve: 4x - 512x - 12 = 4 - 71x + 12
75. Ethan has $60,000 to invest. He puts part of the money in
a CD that earns 3% simple interest per year and the rest
in a mutual fund that earns 8% simple interest per year.
How much did he invest in each if his earned interest the
first year was $3700?
‘Are You Prepared?’ Answers
1. y
2.
2 (1, 1)
y
2
(4, 2)
3. 10, - 82, 12, 02
(1, 1)
2 x
(0, 0)
4
x
(1, 1)
3.5 Graphing Techniques: Transformations
OBJECTIVES 1 Graph Functions Using Vertical and Horizontal Shifts (p. 248)
2 Graph Functions Using Compressions and Stretches (p. 251)
3 Graph Functions Using Reflections about the x-Axis and the y-Axis (p. 253)
At this stage, if you were asked to graph any of the functions defined by y = x,
1
3
y = x2, y = x3, y = 2x, y = 2
x, y = , or y = 0 x 0 , your response should be,
x
“Yes, I recognize these functions and know the general shapes of their graphs.”
(If this is not your answer, review the previous section, Figures 32 through 38.)
248
CHAPTER 3 Functions and Their Graphs
Sometimes we are asked to graph a function that is “almost” like one that we
already know how to graph. In this section, we develop techniques for graphing such
functions. Collectively, these techniques are referred to as transformations.
1 Graph Functions Using Vertical and Horizontal Shifts
EX AMPLE 1
Vertical Shift Up
Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 + 3. Find the domain
and range of g.
Solution
Begin by obtaining some points on the graphs of f and g. For example, when x = 0,
then y = f102 = 0 and y = g102 = 3. When x = 1, then y = f 112 = 1 and
y = g112 = 4. Table 7 lists these and a few other points on each graph. Notice that
each y-coordinate of a point on the graph of g is 3 units larger than the y-coordinate
of the corresponding point on the graph of f . We conclude that the graph of g is
identical to that of f, except that it is shifted vertically up 3 units. See Figure 43.
y = x2 + 3 y
(2, 7)
(2, 7)
Table 7
x
y = f (x)
= x2
y = g (x)
= x2 + 3
-2
4
7
-1
1
4
0
0
3
1
1
4
2
4
7
(1, 4)
(1, 4)
5
(2, 4)
Up 3
units
(2, 4)
(0, 3) y = x 2
(1, 1)
(1, 1)
3
(0, 0)
3
x
Figure 43
The domain of g is all real numbers, or 1 - q , q 2 . The range of g is 3 3, q 2 .
EX AMPLE 2
r
Vertical Shift Down
Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 - 4. Find the domain
and range of g.
Solution
Table 8 lists some points on the graphs of f and g. Notice that each y-coordinate of g
is 4 units less than the corresponding y-coordinate of f.
To obtain the graph of g from the graph of f, subtract 4 from each y-coordinate
on the graph of f. The graph of g is identical to that of f, except that it is shifted
down 4 units. See Figure 44.
y
(– 2, 4)
Table 8
x
y = f (x)
= x2
y = g (x)
= x2 − 4
-2
4
0
-1
1
-3
0
0
-4
1
1
-3
2
4
0
y = x2
4
(2, 4)
Down
4 units
Down
4 units
(2, 0) (0, 0)
(2, 0) 4 x
y = x2 4
5
(0, 4)
Figure 44
The domain of g is all real numbers, or 1 - q , q 2 . The range of g is 3 - 4, q 2 .
r
Note that a vertical shift affects only the range of a function, not the domain. For
example, the range of f(x) = x2 is 3 0, q ). In Example 1 the range of g is 3 3, q ),
whereas in Example 2 the range of g is 3 - 4, q ). The domain for all three functions
is all real numbers.
SECTION 3.5 Graphing Techniques: Transformations
Exploration
Y2 = x 2 + 2
249
On the same screen, graph each of the following functions:
Y1 = x2
6
Y2 = x2 + 2
Y3 = x2 - 2
Y1 =
x2
−6
6
−2
Figure 45 illustrates the graphs. You should have observed a general pattern. With Y1 = x2 on the
screen, the graph of Y2 = x2 + 2 is identical to that of Y1 = x2, except that it is shifted vertically up
2 units. The graph of Y3 = x2 - 2 is identical to that of Y1 = x2, except that it is shifted vertically down
2 units.
Y3 = x 2 − 2
Figure 45
We are led to the following conclusions:
In Words
For y = f (x) + k, k 7 0, add k
to each y-coordinate on the graph
of y = f (x) to shift the graph
up k units.
For y = f (x) - k, k 7 0, subtract
k from each y-coordinate to shift
the graph down k units.
If a positive real number k is added to the output of a function y = f 1x2 , the
graph of the new function y = f 1x2 + k is the graph of f shifted vertically up
k units.
If a positive real number k is subtracted from the output of a function y = f1x2,
the graph of the new function y = f1x2 - k is the graph of f shifted vertically
down k units.
Now Work
EXAMPL E 3
PROBLEM
39
Horizontal Shift to the Right
Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x - 2. Find the
domain and range of g.
Solution
Table 9
The function g1x2 = 1x - 2 is basically a square root function. Table 9 lists some
points on the graphs of f and g. Note that when f1x2 = 0, then x = 0, and when
g1x2 = 0, then x = 2. Also, when f1x2 = 2, then x = 4, and when g1x2 = 2,
then x = 6. Notice that the x-coordinates on the graph of g are 2 units larger than
the corresponding x-coordinates on the graph of f for any given y-coordinate.
We conclude that the graph of g is identical to that of f, except that it is shifted
horizontally 2 units to the right. See Figure 46.
x
y = f (x)
= 1x
x
y = g (x)
= 1x − 2
0
0
2
0
1
1
3
1
4
2
6
2
9
3
11
3
y
5
y = "x
Right 2 units
(4, 2)
y = "x − 2
(6, 2)
(0, 0)
(2, 0)
9
x
Right 2 units
Figure 46
The domain of g is [2, q ) and the range is [0, q ).
EXAMPL E 4
r
Horizontal Shift to the Left
Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x + 4. Find the
domain and range of g.
Solution
The function g1x2 = 1x + 4 is basically a square root function. Its graph is the same as
that of f, except that it is shifted horizontally 4 units to the left. See Figure 47 on page 250.
250
CHAPTER 3 Functions and Their Graphs
y
5
y = "x + 4
Left 4 units
(0, 2)
(4, 2)
(−4, 0)
−5
Figure 47
(0, 0)
y = "x
5
x
Left 4 units
The domain of g is [- 4, q ) and the range is [0, q ).
Now Work
PROBLEM
r
43
Note that a horizontal shift affects only the domain of a function, not the range.
For example, the domain of f1x2 = 1x is [0, q ). In Example 3 the domain of g is
[2, q ), whereas in Example 4 the domain of g is [- 4, q ). The range for all three
functions is [0, q ).
Exploration
Y2 = (x − 3) 2
On the same screen, graph each of the following functions:
Y 1 = x2
Y2 = (x - 3)2
6
Y3 = (x + 2)2
−6
6
Y3 = (x + 2) 2
−2
Y1 = x 2
Figure 48 illustrates the graphs.
You should have observed the following pattern. With the graph of Y1 = x2 on the screen, the
graph of Y2 = (x - 3)2 is identical to that of Y1 = x2, except that it is shifted horizontally to the right
3 units. The graph of Y3 = (x + 2)2 is identical to that of Y1 = x2, except that it is shifted horizontally
to the left 2 units.
Figure 48
We are led to the following conclusions:
If the argument x of a function f is replaced by x - h, h 7 0, the graph of the
new function y = f1x - h2 is the graph of f shifted horizontally right h units.
In Words
For y = f (x - h), h 7 0, add h to
each x-coordinate on the graph of
y = f (x) to shift the graph right
h units. For y = f (x + h), h 7 0,
subtract h from each x-coordinate
on the graph of y = f (x) to shift
the graph left h units.
EX AM PL E 5
If the argument x of a function f is replaced by x + h, h 7 0, the graph of the
new function y = f1x + h2 is the graph of f shifted horizontally left h units.
Observe the distinction between vertical and horizontal shifts. The graph of
f1x2 = x3 + 2 is obtained by shifting the graph of y = x3 up 2 units, because we
evaluate the cube function first and then add 2. The graph of g(x) = 1x + 22 3 is
obtained by shifting the graph of y = x3 left 2 units, because we add 2 to x before
we evaluate the cube function.
Vertical and horizontal shifts are sometimes combined.
Combining Vertical and Horizontal Shifts
Graph the function f1x2 = x + 3 - 5. Find the domain and range of f.
Solution
We graph f in steps. First, note that the rule for f is basically an absolute value function,
so begin with the graph of y = x as shown in Figure 49(a). Next, to get the
graph of y = x + 3 , shift the graph of y = x horizontally 3 units to the left.
See Figure 49(b). Finally, to get the graph of y = x + 3 - 5, shift the graph of
y = x + 3 vertically down 5 units. See Figure 49(c). Note the points plotted on
each graph. Using key points can be helpful in keeping track of the transformation
that has taken place.
251
SECTION 3.5 Graphing Techniques: Transformations
y
y
y
5
5
5
(−1, 2)
(−2, 2)
(2, 2)
(0, 0)
(−5, 2)
5
x
(−3, 0)
x
2
(−1, −3)
(−5, −3)
Replace x by x + 3;
Horizontal shift
left 3 units
y = 0x 0
Subtract 5:
Vertical shift
down 5 units
y = 0 x + 30
(−3, −5)
y = 0 x + 30 − 5
(b)
(a)
x
2
(c)
Figure 49
The domain of f is all real numbers, or 1 - q , q 2. The range of f is [- 5, q ).
Check: Graph Y1 = f 1x2 = x + 3 - 5 and compare the graph to
Figure 49(c).
r
In Example 5, if the vertical shift had been done first, followed by the horizontal
shift, the final graph would have been the same. Try it for yourself.
Now Work
PROBLEMS
45
AND
69
2 Graph Functions Using Compressions and Stretches
EXAMPL E 6
Vertical Stretch
Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 21x.
Solution
Table 10
To see the relationship between the graphs of f and g, we form Table 10, listing points
on each graph. For each x, the y-coordinate of a point on the graph of g is 2 times as
large as the corresponding y-coordinate on the graph of f. The graph of f1x2 = 1x
is vertically stretched by a factor of 2 to obtain the graph of g1x2 = 21x. For
example, 11, 12 is on the graph of f, but 11, 22 is on the graph of g. See Figure 50.
y = 2"x
x
y = f (x)
= 1x
y = g (x)
= 21x
y
0
0
0
5
1
1
2
(4, 4)
4
2
4
(4, 2)
9
3
6
(9, 6)
(1, 2)
(9, 3)
(0, 0)
5
(1, 1)
Vertical Compression
Use the graph of f1x2 = 0 x 0 to obtain the graph of g1x2 =
Solution
10 x
r
Figure 50
EXAMPL E 7
y = "x
1
0x0.
2
1
as large as the
2
corresponding y-coordinate on the graph of f. The graph of f1x2 = 0 x 0 is vertically
1
1
compressed by a factor of to obtain the graph of g1x2 = 0 x 0 . For example, 12, 22 is
2
2
on the graph of f, but 12, 12 is on the graph of g. See Table 11 and Figure 51 on page 252.
For each x, the y-coordinate of a point on the graph of g is
252
CHAPTER 3 Functions and Their Graphs
Table 11
x
In Words
y
y = g (x)
1
= x
2
y = f (x)
= x
-2
2
1
-1
1
1
2
0
0
0
1
1
1
2
2
2
1
y =⏐x⏐
4
y=
(2, 2)
(2, 2)
(– 2, 1)
1– x
⏐ ⏐
2
(2, 1)
4
(0, 0)
4 x
Figure 51
r
When the right side of a function y = f1x2 is multiplied by a positive number a,
the graph of the new function y = af1x2 is obtained by multiplying each
y-coordinate on the graph of y = f1x2 by a. The new graph is a vertically
compressed (if 0 6 a 6 1) or a vertically stretched (if a 7 1) version of the
graph of y = f1x2.
For y = af (x), a 7 0, the
factor a is “outside” the function,
so it affects the y-coordinates.
Multiply each y-coordinate on
the graph of y = f (x) by a.
Now Work
PROBLEM
47
What happens if the argument x of a function y = f 1x2 is multiplied by a
positive number a, creating a new function y = f 1ax2? To find the answer, look at
the following Exploration.
Exploration
On the same screen, graph each of the following functions:
Y1 = f (x) = 1x
Y2 = f (2x) = 12x
1
x
1
Y3 = f a x b =
x =
A2
2
A2
Create a table of values to explore the relation between the x- and y-coordinates of each function.
Result You should have obtained the graphs in Figure 52. Look at Table 12(a). Note that (1, 1), (4, 2),
and (9, 3) are points on the graph of Y1 = 1x. Also, (0.5, 1), (2, 2), and (4.5, 3) are points on the graph of
1
Y2 = 22x. For a given y-coordinate, the x-coordinate on the graph of Y2 is of the x-coordinate on Y1.
2
3
Y2 = √ 2x
Table 12
Y1 = √x
x
Y3 = Ä 2
0
0
Figure 52
4
(a)
(b)
We conclude that the graph of Y2 = 22x is obtained by multiplying the x-coordinate of each point on
1
the graph of Y1 = 1x by . The graph of Y2 = 22x is the graph of Y1 = 1x compressed horizontally.
2
Look at Table 12(b). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also
x
notice that (2, 1), (8, 2), and (18, 3) are points on the graph of Y3 =
. For a given y-coordinate,
A2
the x-coordinate on the graph of Y3 is 2 times the x-coordinate on Y1. We conclude that the graph of
x
Y3 =
is obtained by multiplying the x-coordinate of each point on the graph of Y1 = 1x by 2.
A2
x
The graph of Y3 =
is the graph of Y1 = 1x stretched horizontally.
A2
SECTION 3.5 Graphing Techniques: Transformations
253
Based on the Exploration, we have the following result:
In Words
If the argument x of a function y = f1x2 is multiplied by a positive number a,
then the graph of the new function y = f1ax2 is obtained by multiplying each
1
x-coordinate of y = f1x2 by . A horizontal compression results if a 7 1, and
a
a horizontal stretch results if 0 6 a 6 1.
For y = f (ax), a 7 0, the
factor a is “inside” the function,
so it affects the x-coordinates.
Multiply each x-coordinate on the
1
graph of y = f (x) by .
a
Let’s look at an example.
Graphing Using Stretches and Compressions
EXAMPL E 8
The graph of y = f1x2 is given in Figure 53. Use this graph to find the graphs of
(a) y = 2f 1x2
(b) y = f 13x2
(a) The graph of y = 2f1x2 is obtained by multiplying each y-coordinate of
y = f1x2 by 2. See Figure 54.
Solution
(b) The graph of y = f13x2 is obtained from the graph of y = f 1x2 by multiplying
1
each x-coordinate of y = f 1x2 by . See Figure 55.
3
y
3
y
1
( 2 , 1(
1
3 2 5 3
2
2
3 , 1
2
(
2
3
2
2 52 3
x
1 3
(
(
3 , 2
2
2
PROBLEMS
63(e)
2
x
2
3
Figure 55 y = f 13x2
AND
3
( 2 , 1 (
Figure 54 y = 2f 1x2
Now Work
( 6 , 1( ( 56 , 1(
1
1
Figure 53 y = f 1x2
2
1
x
(
y
( 52 , 2(
2
( 52 , 1(
y f(x)
2
( 2 , 2(
r
(g)
3 Graph Functions Using Reflections about the x-Axis
and the y-Axis
Reflection about the x-Axis
EXAMPL E 9
Graph the function f1x2 = - x2. Find the domain and range of f.
y
(2, 4)
4
(1, 1)
4
point (x, y) on the graph of y = x2, the point 1x, - y2 is on the graph of y = - x2, as
indicated in Table 13. Draw the graph of y = - x2 by reflecting the graph of y = x2
about the x-axis. See Figure 56.
(2, 4)
Table 13
(1, 1)
(1, – 1)
(–1, –1)
(–2, –4)
Solution Begin with the graph of y = x2, as shown in black in Figure 56. For each
y = x2
–4
(2, – 4)
4
x
x
y = x2
y = −x 2
-2
4
-4
-1
1
-1
0
0
0
1
1
-1
2
4
-4
y = –x2
Figure 56
The domain of f is all real numbers, or ( - q , q ). The range of f is ( - q , 0].
r
254
CHAPTER 3 Functions and Their Graphs
When the right side of the function y = f1x2 is multiplied by - 1, the graph of
the new function y = - f 1x2 is the reflection about the x-axis of the graph of
the function y = f 1x2.
Now Work
EX AM P L E 1 0
PROBLEM
49
Reflection about the y-Axis
Graph the function f1x2 = 1- x. Find the domain and range of f.
Solution
To get the graph of f 1x2 = 1- x, begin with the graph of y = 1x, as shown in
Figure 57. For each point 1x, y2 on the graph of y = 1x, the point 1 - x, y2 is on
the graph of y = 1- x. Obtain the graph of y = 1- x by reflecting the graph of
y = 1x about the y-axis. See Figure 57.
y
4
y=
y=
–x
( – 1, 1)
–5
x
(4, 2)
( – 4, 2)
(1, 1)
(0, 0)
5
x
Figure 57
In Words
For y = - f (x), multiply each
y-coordinate on the graph of
y = f (x) by - 1.
For y = f (- x), multiply each
x-coordinate by - 1.
The domain of f is ( - q , 0]. The range of f is the set of all nonnegative real numbers,
or [0, q ).
r
When the graph of the function y = f1x2 is known, the graph of the new
function y = f1 - x2 is the reflection about the y-axis of the graph of the
function y = f1x2.
SUMMARY OF GRAPHING TECHNIQUES
To Graph:
Draw the Graph of f and:
Functional Change to f(x)
Vertical shifts
y = f 1x2 + k, k 7 0
y = f1x2 - k, k 7 0
Raise the graph of f by k units.
Lower the graph of f by k units.
Add k to f1x2.
Subtract k from f1x2.
Horizontal shifts
y = f1x + h2, h 7 0
y = f1x - h2 , h 7 0
Shift the graph of f to the left h units.
Shift the graph of f to the right h units.
Replace x by x + h.
Replace x by x - h.
Multiply each y-coordinate of y = f1x2 by a.
Stretch the graph of f vertically if a 7 1.
Compress the graph of f vertically if 0 6 a 6 1.
1
Multiply each x-coordinate of y = f1x2 by .
a
Stretch the graph of f horizontally if 0 6 a 6 1.
Compress the graph of f horizontally if a 7 1.
Multiply f1x2 by a.
Reflection about the x-axis
y = - f1x2
Reflect the graph of f about the x-axis.
Multiply f1x2 by - 1.
Reflection about the y-axis
y = f 1 - x2
Reflect the graph of f about the y-axis.
Replace x by - x.
Compressing or stretching
y = af 1x2, a 7 0
y = f 1ax2, a 7 0
Replace x by ax.
SECTION 3.5 Graphing Techniques: Transformations
255
Determining the Function Obtained from a Series
of Transformations
EX AM PL E 11
Find the function that is finally graphed after the following three transformations
are applied to the graph of y = 0 x 0 .
1. Shift left 2 units
2. Shift up 3 units
3. Reflect about the y-axis
Solution
y = 0x + 20
1. Shift left 2 units: Replace x by x + 2.
y = 0x + 20 + 3
2. Shift up 3 units: Add 3.
y = 0 -x + 20 + 3
3. Reflect about the y-axis: Replace x by - x.
Now Work
PROBLEM
r
27
Combining Graphing Procedures
EX AM PL E 1 2
3
+ 1. Find the domain and range of f .
x - 2
1
It is helpful to write f as f (x) = 3 a
b + 1. Now use the following steps to
x - 2
obtain the graph of f .
Graph the function f1x2 =
Solution
1
x
STEP 1: y =
Reciprocal function
1
3
STEP 2: y = 3 # a b =
x
x
3
x - 2
3
STEP 4: y =
+ 1
x - 2
STEP 3: y =
Multiply by 3; vertical stretch by a factor of 3.
Replace x by x - 2; horizontal shift to the right 2 units.
Add 1; vertical shift up 1 unit.
See Figure 58.
y
4
y
4
(1, 1)
(1, 3)
4 x
y
4
(3, 3)
3
2, –
2
( )
1
2, –
2
( )
24
y
4
24
3
4, –
2
( )
4 x
4
(21, 21)
x
(4, 5–2 )
4 x
24
(1, 22)
(21, 23)
24
1
x
(a) y 5 ––
Figure 58
(3, 4)
Multiply by 3;
Vertical stretch
(1, 23)
24
24
24
3
x
(b) y 5 ––
Replace x by x 2 2;
Horizontal shift
right 2 units
3
x –2
(c) y 5 –––
Add 1;
Vertical shift
up 1 unit
3
x –2
(d) y 5 ––– 1 1
1
The domain of y = is 5 x x ≠ 06 and its range is 5 y y ≠ 06 . Because we
x
shifted right 2 units and up 1 unit to obtain f , the domain of f is 5 x x ≠ 26 and its
range is 5 y y ≠ 16 .
r
256
CHAPTER 3 Functions and Their Graphs
HINT: Although the order in which
transformations are performed can be
altered, consider using the following
order for consistency:
1. Reflections
2. Compressions and stretches
3. Shifts
Other orderings of the steps shown in Example 12 would also result in the graph
of f. For example, try this one:
STEP 1: y =
1
x
Reciprocal function
1
Replace x by x - 2; horizontal shift to the right 2 units.
x - 2
3
STEP 3: y =
Multiply by 3; vertical stretch by a factor of 3.
x - 2
3
STEP 4: y =
+ 1 Add 1; vertical shift up 1 unit.
x - 2
STEP 2: y =
Combining Graphing Procedures
EX A MPL E 13
Graph the function f1x2 = 21 - x + 2. Find the domain and range of f .
Solution
Because horizontal shifts require the form x - h, begin by rewriting f1x2 as
f1x2 = 21 - x + 2 = 2 - (x - 1) + 2. Now use the following steps.
STEP 1: y = 1 x
Square root function
STEP 2: y = 2 - x
Replace x by - x; reflect about the y-axis.
STEP 3: y = 2 - (x - 1) = 21 - x Replace x by x - 1; horizontal shift to the right 1 unit.
STEP 4: y = 21 - x + 2
Add 2; vertical shift up 2 units.
See Figure 59.
(1, 1)
25
y
5
y
5
y
5
(4, 2)
(a) y 5
(0, 3)
(23, 2)
(1, 2)
(0, 1)
5 x 25
(0, 0)
(24, 2)
y
(23, 4) 5
x Replace x by 2x;
Reflect
about y-axis
(21, 1)
(0, 0)
(b) y 5 2x
5 x 25
(1, 0)
Replace x by x 2 1; (c) y 5
Horizontal shift
5
right 1 unit
5
5 x 25
2(x 2 1) Add 2;
(d) y 5
2 x 1 1 Vertical shift
up 2 units
12x
Figure 59
The domain of f is ( - q , 1] and the range is [2, q ).
Now Work
PROBLEM
5 x
12x12
r
55
3.5 Assess Your Understanding
Concepts and Vocabulary
1. Suppose that the graph of a function f is known. Then the
graph of y = f1x - 22 may be obtained by a(n)
shift of the graph of f to the
a distance of 2 units.
2. Suppose that the graph of a function f is known. Then the
graph of y = f 1 - x2 may be obtained by a reflection about
the
-axis of the graph of the function y = f 1x2.
1
3. True or False The graph of y = g(x) is the graph of
3
y = g(x) vertically stretched by a factor of 3.
4. True or False The graph of y = - f 1x2 is the reflection
about the x-axis of the graph of y = f 1x2.
5. Which of the following functions has a graph that is the
graph of y = 2x shifted down 3 units?
(a) y = 2x + 3
(b) y = 2x - 3
(c) y = 2x + 3
(d) y = 2x - 3
6. Which of the following functions has a graph that is the
graph of y = f (x) compressed horizontally by a factor of 4?
1
(a) y = f (4x)
(b) y = f ¢ x≤
4
(c) y = 4f (x)
(d) y =
1
f (x)
4
257
SECTION 3.5 Graphing Techniques: Transformations
Skill Building
In Problems 7–18, match each graph to one of the following functions:
A. y = x2 + 2
E. y = 1x - 22
I. y = 2x
7.
C. y = 0 x 0 + 2
B. y = - x2 + 2
2
F. y = - 1x + 22
J. y = - 2x
2
y
3
H. y = - 0 x + 2 0
K. y = 2 0 x 0
2
8.
D. y = - 0 x 0 + 2
G. y = 0 x - 2 0
2
L. y = - 2 0 x 0
9.
y
3
10.
y
1
3
3 x
3
11.
12.
y
3
3
3x
3
3 x
3
13.
y
5
y
3
3 x
14.
y
3
3
3 x
y
8
3 x
6
3
3
15.
16.
y
4
4
4 x
1
3 x
3
3 x
18.
y
4
4
3
4
4
3
17.
y
3
6 x
4 x
y
3
3
4
3 x
3
In Problems 19–26, write the function whose graph is the graph of y = x3, but is:
19. Shifted to the right 4 units
20. Shifted to the left 4 units
21. Shifted up 4 units
22. Shifted down 4 units
23. Reflected about the y-axis
24. Reflected about the x-axis
25. Vertically stretched by a factor of 4
26. Horizontally stretched by a factor of 4
In Problems 27–30, find the function that is finally graphed after each of the following transformations is applied to the graph of
y = 1x in the order stated.
27. (1) Shift up 2 units
(2) Reflect about the x-axis
(3) Reflect about the y-axis
28. (1) Reflect about the x-axis
(2) Shift right 3 units
(3) Shift down 2 units
29. (1) Reflect about the x-axis
(2) Shift up 2 units
(3) Shift left 3 units
30. (1) Shift up 2 units
(2) Reflect about the y-axis
(3) Shift left 3 units
31. If 13, 62 is a point on the graph of y = f 1x2, which of the
following points must be on the graph of y = - f 1x2?
(a) 16, 32
(b) 16, - 32
32. If 13, 62 is a point on the graph of y = f 1x2, which of the
following points must be on the graph of y = f 1 - x2?
(a) 16, 32
(b) 16, - 32
33. If 11, 32 is a point on the graph of y = f 1x2, which of the
following points must be on the graph of y = 2f 1x2?
3
(a) a1, b
(b) 12, 32
2
1
(c) 11, 62
(d) a , 3b
2
34. If 14, 22 is a point on the graph of y = f 1x2, which of the
following points must be on the graph of y = f 12x2 ?
(a) 14, 12
(b) 18, 22
(c) 13, - 62
(d) 1 - 3, 62
(c) 13, - 62
(d) 1 - 3, 62
(c) 12, 22
(d) 14, 42
258
CHAPTER 3 Functions and Their Graphs
35. Suppose that the x-intercepts of the graph of y = f 1x2
are - 5 and 3.
(a) What are the x-intercepts of the graph of y = f 1x + 22?
(b) What are the x-intercepts of the graph of y = f 1x - 22?
(c) What are the x-intercepts of the graph of y = 4f 1x2?
(d) What are the x-intercepts of the graph of y = f 1 - x2?
36. Suppose that the x-intercepts of the graph of y = f 1x2
are - 8 and 1.
(a) What are the x-intercepts of the graph of y = f 1x + 42?
(b) What are the x-intercepts of the graph of y = f 1x - 32?
(c) What are the x-intercepts of the graph of y = 2f 1x2?
(d) What are the x-intercepts of the graph of y = f 1 - x2?
37. Suppose that the function y = f 1x2 is increasing on the
interval 1 - 1, 52.
(a) Over what interval is the graph of y = f 1x + 22
increasing?
(b) Over what interval is the graph of y = f 1x - 52
increasing?
(c) What can be said about the graph of y = - f 1x2?
(d) What can be said about the graph of y = f 1 - x2?
38. Suppose that the function y = f 1x2 is decreasing on the
interval 1 - 2, 72.
(a) Over what interval is the graph of y = f 1x + 22
decreasing?
(b) Over what interval is the graph of y = f 1x - 52
decreasing?
(c) What can be said about the graph of y = - f 1x2?
(d) What can be said about the graph of y = f 1 - x2?
In Problems 39–62, graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph
of the basic function (for example, y = x2) and show all stages. Be sure to show at least three key points. Find the domain and the range
of each function.
39. f 1x2 = x2 - 1
40. f 1x2 = x2 + 4
41. g1x2 = x3 + 1
42. g1x2 = x3 - 1
43. h1x2 = 2x + 2
44. h1x2 = 2x + 1
45. f 1x2 = 1x - 12 3 + 2
46. f 1x2 = 1x + 22 3 - 3
47. g1x2 = 41x
3
49. f 1x2 = - 2
x
50. f 1x2 = - 1x
51. f 1x2 = 21x + 12 2 - 3
52. f 1x2 = 31x - 22 2 + 1
53. g1x2 = 22x - 2 + 1
54. g1x2 = 3 0 x + 1 0 - 3
55. h1x2 = 1 - x - 2
56. h1x2 =
57. f 1x2 = - 1x + 12 3 - 1
58. f 1x2 = - 42x - 1
59. g1x2 = 2 0 1 - x 0
60. g1x2 = 422 - x
61. h1x2 =
48. g1x2 =
1
1x
2
1
2x
4
+ 2
x
3
62. h1x2 = 2x - 1 + 3
In Problems 63–66, the graph of a function f is illustrated. Use the graph of f as the first step toward graphing each of the following
functions:
(a) F 1x2 = f 1x2 + 3
(b) G1x2 = f 1x + 22
(c) P 1x2 = - f 1x2
1
(e) Q1x2 = f 1x2
2
(f) g1x2 = f 1 - x2
(g) h1x2 = f 12x2
y
4
(0, 2)
63.
y
4
64.
(2, 2)
2
1
(2, 2)
(4, 0)
4
(4, 2)
2
x
2
y
65.
4 2
(4, 2)
2
(2, 2)
4 x
2
(4, 2)
π
π
–2
1
π
( –2 , 1)
(d) H1x2 = f 1x + 12 - 2
π–
2
y
66.
(π–2 , 1)
1
π
x
π
π
–2
(π, 1)
π
π–
1
2
x
(π, 1)
Mixed Practice
In Problems 67–74, complete the square of each quadratic expression. Then graph each function using the technique of shifting.
(If necessary, refer to Chapter R, Section R.5 to review completing the square.)
67. f 1x2 = x2 + 2x
71. f 1x2 = 2x2 - 12x + 19
68. f 1x2 = x2 - 6x
72. f 1x2 = 3x2 + 6x + 1
69. f 1x2 = x2 - 8x + 1
73. f 1x2 = - 3x2 - 12x - 17
70. f 1x2 = x2 + 4x + 2
74. f 1x2 = - 2x2 - 12x - 13
SECTION 3.5 Graphing Techniques: Transformations
259
Applications and Extensions
75. The graph of a function f is illustrated in the figure.
(a) Draw the graph of y = 0 f 1x2 0 .
y
2
(1, 1)
3
(2, 1)
(2, 0) 3 x
(1, 1)
2
(b) Draw the graph of y = f 1 0 x 0 2.
72
(2, 0)
23
(21, 72)
68
64
(6, 65)
60
56
t
4 8 12 16 20 24
Time (hours after midnight)
84. Digital Music Revenues The total projected worldwide
digital music revenues R, in millions of dollars, for the years
2012 through 2017 can be estimated by the function
R 1x2 = 28.6x2 + 300x + 4843
(1, 1)
(22, 0)
3 x
(21, 21)
22
76
0
76. The graph of a function f is illustrated in the figure.
(a) Draw the graph of y = 0 f 1x2 0 .
y
2
80
Temperature (°) )
(b) Draw the graph of y = f 1 0 x 0 2.
T
(0, 21)
77. Suppose 11, 32 is a point on the graph of y = f 1x2 .
(a) What point is on the graph of y = f 1x + 32 - 5?
(b) What point is on the graph of y = - 2f 1x - 22 + 1?
(c) What point is on the graph of y = f 12x + 32?
78. Suppose 1 - 3, 52 is a point on the graph of y = g1x2 .
(a) What point is on the graph of y = g1x + 12 - 3?
(b) What point is on the graph of y = - 3g1x - 42 + 3?
(c) What point is on the graph of y = g13x + 92?
79. Graph the following functions using transformations.
(b) g(x) = - int(x)
(a) f(x) = int( - x)
80. Graph the following functions using transformations
(a) f(x) = int(x - 1)
(b) g(x) = int(1 - x)
81. (a) Graphf(x) = x - 3 - 3 using transformations.
(b) Find the area of the region that is bounded by f and the
x-axis and lies below the x-axis.
82. (a) Graph f(x) = - 2 x - 4 + 4 using transformations.
(b) Find the area of the region that is bounded by f and the
x-axis and lies above the x-axis.
83. Thermostat Control Energy conservation experts estimate
that homeowners can save 5% to 10% on winter heating
bills by programming their thermostats 5 to 10 degrees lower
while sleeping. In the graph (top, right), the temperature T
(in degrees Fahrenheit) of a home is given as a function of
time t (in hours after midnight) over a 24-hour period.
(a) At what temperature is the thermostat set during
daytime hours? At what temperature is the thermostat
set overnight?
(b) The
homeowner
reprograms
the
thermostat
to y = T 1t2 - 2. Explain how this affects the
temperature in the house. Graph this new function.
(c) The
homeowner
reprograms
the
thermostat
to y = T 1t + 12. Explain how this affects the
temperature in the house. Graph this new function.
Source: Roger Albright, 547 Ways to Be Fuel Smart, 2000
where x is the number of years after 2012.
(a) Find R 102, R 132, and R 152 and explain what each
value represents.
(b) Find r 1x2 = R 1x - 22.
(c) Find r 122, r 152, and r 172 and explain what each value
represents.
(d) In the model r = r 1x2 , what does x represent?
(e) Would there be an advantage in using the model r when
estimating the projected revenues for a given year
instead of the model R ?
Source: IFPI Digital Music Report 2013
85. Temperature Measurements The relationship between
the Celsius (°C) and Fahrenheit (°F) scales for measuring
temperature is given by the equation
F =
9
C + 32
5
The relationship between the Celsius (°C) and Kelvin (K)
9
scales is K = C + 273. Graph the equation F = C + 32
5
using degrees Fahrenheit on the y-axis and degrees Celsius
on the x-axis. Use the techniques introduced in this section
to obtain the graph showing the relationship between Kelvin
and Fahrenheit temperatures.
86. Period of a Pendulum The period T (in seconds) of a simple
pendulum is a function of its length l (in feet) defined by the
equation
T = 2p
l
Ag
where g ≈ 32.2 feet per second per second is the acceleration
due to gravity. (Continued on page 260.)
260
CHAPTER 3 Functions and Their Graphs
(a) Use a graphing utility to graph the function T = T 1l2.
(b) Now graph the functions T = T 1l + 12, T = T 1l + 22,
and T = T 1l + 32.
(c) Discuss how adding to the length l changes the period T.
(d) Now graph the functions T = T 12l2, T = T 13l2, and
T = T 14l2.
(e) Discuss how multiplying the length l by factors of 2, 3,
and 4 changes the period T.
87. The equation y = 1x - c2 2 defines a family of parabolas,
one parabola for each value of c. On one set of coordinate
axes, graph the members of the family for c = 0, c = 3, and
c = - 2.
88. Repeat Problem 87 for the family of parabolas y = x2 + c.
Explaining Concepts: Discussion and Writing
89. Suppose that the graph of a function f is known. Explain
how the graph of y = 4f 1x2 differs from the graph of
y = f 14x2.
ideas presented in this section, what do you think is the area
under the curve of y = 1- x bounded from below by the
x-axis and on the left by x = - 4? Justify your answer.
90. Suppose that the graph of a function f is known. Explain
how the graph of y = f 1x2 - 2 differs from the graph of
y = f 1x - 22.
92. Explain how the range of the function f (x) = x2 compares
to the range of g(x) = f (x) + k.
91. The area under the curve y = 1x bounded from below by the
16
square units. Using the
x-axis and on the right by x = 4 is
3
93. Explain how the domain of g(x) = 2x compares to the
domain of g(x - k), where k Ú 0.
Retain Your Knowledge
Problems 94–97 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
94. Determine the slope and y-intercept of the graph of 3x - 5y = 30.
1x -2y3 2 4
1x2y -5 2 -2
96. The amount of water used when taking a shower varies directly with the number of minutes the shower is run. If a 4-minute
shower uses 7 gallons of water, how much water is used in a 9-minute shower?
95. Simplify
97. List the intercepts and test for symmetry: y2 = x + 4
3.6 Mathematical Models: Building Functions
OBJECTIVE 1 Build and Analyze Functions (p. 260)
1 Build and Analyze Functions
Real-world problems often result in mathematical models that involve functions.
These functions need to be constructed or built based on the information given.
In building functions, we must be able to translate the verbal description into the
language of mathematics. This is done by assigning symbols to represent the
independent and dependent variables and then by finding the function or rule that
relates these variables.
EX AMPLE 1
Finding the Distance from the Origin to a Point on a Graph
Let P = 1x, y2 be a point on the graph of y = x2 - 1.
(a) Express the distance d from P to the origin O as a function of x.
(b) What is d if x = 0?
(c) What is d if x = 1?
22
?
(d) What is d if x =
2
SECTION 3.6 Mathematical Models: Building Functions
261
(e) Use a graphing utility to graph the function d = d 1x2, x Ú 0. Rounding to two
decimal places, find the value(s) of x at which d has a local minimum. [This gives
the point(s) on the graph of y = x2 - 1 closest to the origin.]
Solution
y
(a) Figure 60 illustrates the graph of y = x2 - 1. The distance d from P to O is
2
d = 2 1x - 02 2 + 1y - 02 2 = 2x2 + y2
1
P5 (x, y)
(0, 0) d
1
21
2 x
Since P is a point on the graph of y = x2 - 1, substitute x2 - 1 for y. Then
d 1x2 = 2x2 + 1x2 - 12 2 = 2x4 - x2 + 1
21
Figure 60 y = x2 - 1
The distance d is expressed as a function of x.
(b) If x = 0, the distance d is
d 102 = 204 - 02 + 1 = 21 = 1
(c) If x = 1, the distance d is
d 112 = 214 - 12 + 1 = 1
(d) If x =
22
, the distance d is
2
da
2
12
12 4
12 2
1
1
13
b =
a
b - a
b + 1 =
- + 1 =
2
B 2
2
B4
2
2
(e) Figure 61 shows the graph of Y1 = 2x4 - x2 + 1. Using the MINIMUM
0
2
Figure 61 d(x) = 2x4 - x2 + 1
feature on a graphing utility, we find that when x ≈ 0.71 the value of d is
smallest. The local minimum is d ≈ 0.87 rounded to two decimal places. Since
d 1x2 is even, it follows by symmetry that when x ≈ - 0.71, the value of d is also
a local minimum. Since 1 {0.712 2 - 1 ≈ - 0.50, the points 1 - 0.71, - 0.502 and
10.71, - 0.502 on the graph of y = x2 - 1 are closest to the origin.
r
Now Work
EXAMPL E 2
y
30
20
1
y 25 x 2
1 2 3 4 5
x
(0,0)
Figure 62
1
Area of a Rectangle
A rectangle has one corner in quadrant I on the graph of y = 25 - x2, another at
the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See
Figure 62.
(x, y)
10
PROBLEM
Solution
(a) Express the area A of the rectangle as a function of x.
(b) What is the domain of A?
(c) Graph A = A 1x2.
(d) For what value of x is the area largest?
(a) The area A of the rectangle is A = xy, where y = 25 - x2. Substituting this
expression for y, we obtain A 1x2 = x125 - x2 2 = 25x - x3.
(b) Since 1x, y2 is in quadrant I, we have x 7 0. Also, y = 25 - x2 7 0, which
implies that x2 6 25, so - 5 6 x 6 5. Combining these restrictions, we have the
domain of A as 5 x 0 6 x 6 56 , or 10, 52 using interval notation.
(c) See Figure 63 on page 262 for the graph of A = A 1x2.
(d) Using MAXIMUM, we find that the maximum area is 48.11 square units at
x = 2.89 units, each rounded to two decimal places. See Figure 64.
262
CHAPTER 3 Functions and Their Graphs
50
50
5
0
0
Figure 63 A(x) = 25x - x3
Now Work
PROBLEM
5
0
0
Figure 64
r
7
Close Call?
EX AMPLE 3
Suppose two planes flying at the same altitude are headed toward each other. One
plane is flying due south at a groundspeed of 400 miles per hour and is 600 miles
from the potential intersection point of the planes. The other plane is flying due
west with a groundspeed of 250 miles per hour and is 400 miles from the potential
intersection point of the planes. See Figure 65.
(a) Build a model that expresses the distance d between the planes as a function of
time t.
(b) Use a graphing utility to graph d = d(t). How close do the planes come to each
other? At what time are the planes closest?
Solution
N
Plane
600 miles
400 mph
(a) Refer to Figure 65. The distance d between the two planes is the hypotenuse of
a right triangle. At any time t, the length of the north/south leg of the triangle
is 600 - 400t. At any time t, the length of the east/west leg of the triangle is
400 - 250t. Use the Pythagorean Theorem to find that the square of the
distance between the two planes is
d
d 2 = (600 - 400t)2 + (400 - 250t)2
Plane
250 mph
E
400 miles
Therefore, the distance between the two planes as a function of time is given by
the model
d(t) = 2(600 - 400t)2 + (400 - 250t)2
Figure 65
(b) Figure 66(a) shows the graph of d = d(t). Using MINIMUM, the minimum
distance between the planes is 21.20 miles, and the time at which the planes are
closest is after 1.53 hours, each rounded to two decimal places. See Figure 66(b).
500
2
0
−50
Figure 66
(b)
(a)
Now Work
PROBLEM
19
r
SECTION 3.6 Mathematical Models: Building Functions
263
3.6 Assess Your Understanding
Applications and Extensions
1. Let P = 1x, y2 be a point on the graph of y = x2 - 8.
(a) Express the distance d from P to the origin as a function
of x.
(b) What is d if x = 0?
(c) What is d if x = 1?
(d) Use a graphing utility to graph d = d1x2.
(e) For what values of x is d smallest?
2. Let P = 1x, y2 be a point on the graph of y = x2 - 8.
(a) Express the distance d from P to the point 10, - 12 as a
function of x.
(b) What is d if x = 0?
(c) What is d if x = - 1?
(d) Use a graphing utility to graph d = d1x2.
(e) For what values of x is d smallest?
3. Let P = 1x, y2 be a point on the graph of y = 1x.
(a) Express the distance d from P to the point 11, 02 as a
function of x.
(b) Use a graphing utility to graph d = d1x2.
(c) For what values of x is d smallest?
4. Let P = 1x, y2 be a point on the graph of y =
y
y 16 x 2
16
(x, y)
8
x
4
(0,0)
8. A rectangle is inscribed in a semicircle of radius 2. See the
figure. Let P = 1x, y2 be the point in quadrant I that is a
vertex of the rectangle and is on the circle.
y
y 4 x2
P (x, y )
2
2
x
1
.
x
(a) Express the distance d from P to the origin as a function
of x.
(b) Use a graphing utility to graph d = d1x2.
(c) For what values of x is d smallest?
(a) Express the area A of the rectangle as a function
of x.
(b) Express the perimeter p of the rectangle as a
function of x.
(c) Graph A = A1x2. For what value of x is A largest?
(d) Graph p = p1x2. For what value of x is p largest?
5. A right triangle has one vertex on the graph of y = x3, x 7 0,
at 1x, y2, another at the origin, and the third on the positive
y-axis at 10, y2, as shown in the figure. Express the area A of
the triangle as a function of x.
9. A rectangle is inscribed in a circle of radius 2. See the figure.
Let P = 1x, y2 be the point in quadrant I that is a vertex of
the rectangle and is on the circle.
y
2
y
P (x, y)
y x3
2
(0, y)
(x, y)
2
x
2
2
2
x y 4
(0, 0)
x
6. A right triangle has one vertex on the graph of
y = 9 - x2, x 7 0, at 1x, y2, another at the origin, and the
third on the positive x-axis at 1x, 02. Express the area A of
the triangle as a function of x.
7. A rectangle has one corner in quadrant I on the graph of
y = 16 - x2, another at the origin, a third on the positive
y-axis, and the fourth on the positive x-axis. See the figure
(top, right).
(a) Express the area A of the rectangle as a function
of x.
(b) What is the domain of A?
(c) Graph A = A1x2. For what value of x is A largest?
(a) Express the area A of the rectangle as a function
of x.
(b) Express the perimeter p of the rectangle as a
function of x.
(c) Graph A = A1x2. For what value of x is A largest?
(d) Graph p = p1x2. For what value of x is p largest?
10. A circle of radius r is inscribed in a square. See the figure.
r
(a) Express the area A of the square as a function of the
radius r of the circle.
(b) Express the perimeter p of the square as a function
of r.
264
CHAPTER 3 Functions and Their Graphs
11. Geometry A wire 10 meters long is to be cut into two pieces.
One piece will be shaped as a square, and the other piece
will be shaped as a circle. See the figure.
40 miles per hour (see the figure). Build a model that
expresses the distance d between the cars as a function of
the time t.
[Hint: At t = 0, the cars leave the intersection.]
x
4x
N
W
10 m
10 4x
E
S
(a) Express the total area A enclosed by the pieces of wire
as a function of the length x of a side of the square.
(b) What is the domain of A ?
(c) Graph A = A1x2. For what value of x is A smallest?
12. Geometry A wire 10 meters long is to be cut into two pieces.
One piece will be shaped as an equilateral triangle, and the
other piece will be shaped as a circle.
(a) Express the total area A enclosed by the pieces of wire
as a function of the length x of a side of the equilateral
triangle.
(b) What is the domain of A ?
(c) Graph A = A1x2. For what value of x is A smallest?
13. Geometry A wire of length x is bent into the shape of a circle.
(a) Express the circumference C of the circle as a function
of x.
(b) Express the area A of the circle as a function of x.
14. Geometry A wire of length x is bent into the shape of a square.
(a) Express the perimeter p of the square as a function of
x.
(b) Express the area A of the square as a function of x.
15. Geometry A semicircle of radius r is inscribed in a rectangle
so that the diameter of the semicircle is the length of the
rectangle. See the figure.
d
19. Uniform Motion Two cars are approaching an intersection.
One is 2 miles south of the intersection and is moving at a
constant speed of 30 miles per hour. At the same time, the
other car is 3 miles east of the intersection and is moving at
a constant speed of 40 miles per hour.
(a) Build a model that expresses the distance d between the
cars as a function of time t.
[Hint: At t = 0, the cars are 2 miles south and 3 miles
east of the intersection, respectively.]
(b) Use a graphing utility to graph d = d1t2. For what
value of t is d smallest?
20. Inscribing a Cylinder in a Sphere Inscribe a right circular
cylinder of height h and radius r in a sphere of fixed radius R.
See the illustration. Express the volume V of the cylinder as
a function of h.
[Hint: V = pr 2 h. Note also the right triangle.]
r
r
(a) Express the area A of the rectangle as a function of the
radius r of the semicircle.
(b) Express the perimeter p of the rectangle as a function
of r.
16. Geometry An equilateral triangle is inscribed in a circle of
radius r. See the figure. Express the circumference C of the
circle as a function of the length x of a side of the triangle.
[Hint: First show that r 2 =
h
Sphere
21. Inscribing a Cylinder in a Cone Inscribe a right circular
cylinder of height h and radius r in a cone of fixed radius R
and fixed height H. See the illustration. Express the volume V
of the cylinder as a function of r.
x2
.]
3
x
R
x
[Hint: V = pr 2 h. Note also the similar triangles.]
r
r
x
17. Geometry An equilateral triangle is inscribed in a circle of
radius r. See the figure in Problem 16. Express the area A
within the circle, but outside the triangle, as a function of the
length x of a side of the triangle.
18. Uniform Motion Two cars leave an intersection at the same
time. One is headed south at a constant speed of 30 miles
per hour, and the other is headed west at a constant speed of
H
h
R
Cone
SECTION 3.6 Mathematical Models: Building Functions
22. Installing Cable TV MetroMedia Cable is asked to
provide service to a customer whose house is located 2 miles
from the road along which the cable is buried. The nearest
connection box for the cable is located 5 miles down the
road. See the figure.
265
24. Filling a Conical Tank Water is poured into a container
in the shape of a right circular cone with radius 4 feet and
height 16 feet. See the figure. Express the volume V of the
water in the cone as a function of the height h of the water.
[Hint: The volume V of a cone of radius r and height h is
1
V = pr 2 h.]
3
House
Stream
4
2 mi
Box
5 mi
h
(a) If the installation cost is $500 per mile along the road
and $700 per mile off the road, build a model that
expresses the total cost C of installation as a function of
the distance x (in miles) from the connection box to
the point where the cable installation turns off the road.
Find the domain of C = C 1x2.
(b) Compute the cost if x = 1 mile.
(c) Compute the cost if x = 3 miles.
(d) Graph the function C = C 1x2. Use TRACE to see how
the cost C varies as x changes from 0 to 5.
(e) What value of x results in the least cost?
23. Time Required to Go from an Island to a Town An island
is 2 miles from the nearest point P on a straight shoreline. A
town is 12 miles down the shore from P. See the illustration.
d2 Town
P
12 x
2 mi
x
r
16
x
12 mi
d1
Island
(a) If a person can row a boat at an average speed of 3 miles
per hour and the same person can walk 5 miles per hour,
build a model that expresses the time T that it takes to
go from the island to town as a function of the distance x
from P to where the person lands the boat.
(b) What is the domain of T ?
(c) How long will it take to travel from the island to town if
the person lands the boat 4 miles from P?
(d) How long will it take if the person lands the boat 8 miles
from P?
25. Constructing an Open Box An open box with a square base
is to be made from a square piece of cardboard 24 inches on
a side by cutting out a square from each corner and turning
up the sides. See the figure.
x
x
x
x
24 in.
x
x
x
x
24 in.
(a) Express the volume V of the box as a function of
the length x of the side of the square cut from each
corner.
(b) What is the volume if a 3-inch square is cut out?
(c) What is the volume if a 10-inch square is cut out?
(d) Graph V = V 1x2. For what value of x is V largest?
26. Constructing an Open Box An open box with a square base
is required to have a volume of 10 cubic feet.
(a) Express the amount A of material used to make such a
box as a function of the length x of a side of the square
base.
(b) How much material is required for a base 1 foot by
1 foot?
(c) How much material is required for a base 2 feet by
2 feet?
(d) Use a graphing utility to graph A = A1x2. For what
value of x is A smallest?
266
CHAPTER 3 Functions and Their Graphs
Retain Your Knowledge
Problems 27–30 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
27. Solve: 2x - 3 - 5 = - 2
28. A 16-foot long Ford Fusion wants to pass a 50-foot truck traveling at 55 mi/h. How fast must the car travel to completely pass the
truck in 5 seconds?
10
29. Find the slope of the line containing the points (3, - 2) and (1, 6).
30. Find the missing length x for the given pair of similar triangles.
4
x
14
Chapter Review
Library of Functions
Square function (p. 239)
Identity function (p. 239)
Constant function (p. 239)
f 1x2 = b
f 1x2 = x
The graph is a horizontal line with
y-intercept b.
f 1x2 = x2
The graph is a line with slope 1 and
y-intercept 0.
y
y
3
y
f (x ) = b
The graph is a parabola with intercept
at 10, 02.
( – 2, 4)
(2, 4)
4
(0,b)
(1, 1)
x
(– 1, 1)
3 x
(0, 0)
–3
( – 1, – 1)
f 1x2 = x
Square root function (p. 240)
y
y
4
2
(1, 1)
4
(1, 1)
Cube root function (p. 240)
f 1x2 = 1x
3
(0, 0)
4
x
1
4 x
(0, 0)
–4
Cube function (p. 240)
(1, 1)
3
f 1x2 = 2x
y
3
(1, 1)
(4, 2)
5 x
(0, 0)
3
(
1–8 , 1–2
(1, 1)
)
(2, 2 )
( 1–8 , 1–2)
3
3 x
(0, 0)
3
4
(1, 1)
(2, 2 )
3
Absolute value function (p. 240)
Reciprocal function (p. 240)
f 1x2 =
f 1x2 = 0 x 0
1
x
y
y
2
3
2
(1, 1)
2 x
(1, 1)
3
f 1x2 = int 1x2
y
4
(2, 2)
(2, 2)
(1, 1)
Greatest integer function (p. 241)
(0, 0)
2
(1, 1)
3 x
2
2
3
2
4
x
Chapter Review
267
Things to Know
Function (pp. 199–202)
A relation between two sets so that each element x in the first set, the domain, has
corresponding to it exactly one element y in the second set, the range. The range is
the set of y-values of the function for the x-values in the domain.
A function can also be characterized as a set of ordered pairs 1x, y2 in which no first
element is paired with two different second elements.
Function notation (pp. 202–205)
y = f 1x2
f is a symbol for the function.
x is the argument, or independent variable.
y is the dependent variable.
f 1x2 is the value of the function at x, or the image of x.
A function f may be defined implicitly by an equation involving x and y or explicitly
by writing y = f 1x2.
Difference quotient of f (p. 205)
f 1x + h2 - f 1x2
h
h ≠ 0
Domain (pp. 206–208)
If unspecified, the domain of a function f defined by an equation is the largest set of
real numbers for which f 1x2 is a real number.
Vertical-line test (p. 214)
A set of points in the xy-plane is the graph of a function if and only if every vertical
line intersects the graph in at most one point.
Even function f (p. 224)
Odd function f (p. 224)
f 1 - x2 = f 1x2 for every x in the domain ( - x must also be in the domain).
f 1 - x2 = - f 1x2 for every x in the domain ( - x must also be in the domain).
Increasing function (p. 226)
A function f is increasing on an open interval I if, for any choice of x1 and x2 in I,
with x1 6 x2 , we have f 1x1 2 6 f 1x2 2.
Decreasing function (p. 226)
A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I,
with x1 6 x2 , we have f 1x1 2 7 f 1x2 2.
Constant function (p. 226)
A function f is constant on an open interval I if, for all choices of x in I, the values of
f 1x2 are equal.
Local maximum (p. 227)
A function f , defined on some interval I, has a local maximum at c if there is an open
interval in I containing c such that, for all x in this open interval, f 1x2 … f 1c2.
Local minimum (p. 227)
A function f , defined on some interval I, has a local minimum at c if there is an open
interval in I containing c such that, for all x in this open interval, f 1x2 Ú f 1c2.
Absolute maximum and
Absolute minimum (p. 227)
Let f denote a function defined on some interval I.
If there is a number u in I for which f 1x2 … f 1u2 for all x in I, then f has an absolute
maximum at u, and the number f 1u2 is the absolute maximum of f on I.
If there is a number v in I for which f 1x2 Ú f 1v2 , for all x in I, then f has an
absolute minimum at v, and the number f 1v2 is the absolute minimum of f on I.
Average rate of change of a function (p. 230)
The average rate of change of f from a to b is
f 1b2 - f 1a2
∆y
=
∆x
b - a
a ≠ b
Objectives
Section You should be able to . . .
Examples
Review Exercises
3.1
1–5
6, 7
8
9, 10
1, 2
3–5, 39
15
6–11
11
1
12–14
27, 28
2–4
16(a)–(e), 17(a), 17(e), 17(g)
1
2
3
4
5
3.2
1
2
Determine whether a relation represents a function (p. 199)
Find the value of a function (p. 202)
Find the difference quotient of a function (p. 205)
Find the domain of a function defined by an equation (p. 206)
Form the sum, difference, product, and quotient of
two functions (p. 208)
Identify the graph of a function (p. 214)
Obtain information from or about the graph
of a function (p. 215)
268
CHAPTER 3 Functions and Their Graphs
Section
3.3
1
2
3
4
5
6
7
3.4
1
2
3.5
1
2
3
3.6
1
You should be able to . . .
Examples
Review Exercises
Determine even and odd functions from a graph (p. 223)
Identify even and odd functions from an equation (p. 225)
Use a graph to determine where a function is increasing,
decreasing, or constant (p. 225)
Use a graph to locate local maxima and local minima (p. 226)
Use a graph to locate the absolute maximum and the
absolute minimum (p. 227)
Use a graphing utility to approximate local maxima and
local minima and to determine where a function is increasing
or decreasing (p. 229)
Find the average rate of change of a function (p. 230)
Graph the functions listed in the library of functions (p. 237)
Graph piecewise-defined functions (p. 242)
Graph functions using vertical and horizontal shifts (p. 248)
Graph functions using compressions and stretches (p. 251)
Graph functions using reflections about the x-axis and
the y-axis (p. 253)
Build and analyze functions (p. 260)
1
2
17(f)
18–21
3
4
17(b)
17(c)
5
17(d)
6
7, 8
1, 2
3, 4
1–5, 11–13
6–8, 12
22, 23, 40(d), 41(b)
24–26
29, 30
37, 38
16(f), 31, 33–36
16(g), 32, 36
9, 10, 11, 13
1–3
16(h), 32, 34, 36
40, 41
Review Exercises
In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range.
1. 5 1 - 1, 02, 12, 32, 14, 02 6
2. 5 14, - 12, 12, 12, 14, 22 6
In Problems 3–5, find the following for each function:
(a) f 122
(b) f 1 - 22
3x
3. f 1x2 = 2
x - 1
(c) f 1 - x2
(d) - f 1x2
(e) f 1x - 22
(f) f 12x2
5. f 1x2 =
4. f 1x2 = 2x2 - 4
In Problems 6–11, find the domain of each function.
6. f 1x2 =
x
x - 9
9. f 1x2 =
x
x + 2x - 3
7. f 1x2 = 22 - x
2
2
10. f(x) =
In Problems 12–14, find f + g, f - g, f # g, and
12. f 1x2 = 2 - x; g1x2 = 3x + 1
8. g1x2 =
2x + 1
x2 - 4
11. g(x) =
x2 - 4
x2
0x0
x
x
2x + 8
f
for each pair of functions. State the domain of each of these functions.
g
13. f 1x2 = 3x2 + x + 1; g1x2 = 3x
15. Find the difference quotient of f 1x2 = - 2x2 + x + 1; that is, find
16. Consider the graph of the function f on the right.
(a) Find the domain and the range of f.
(b) List the intercepts.
(c) Find f 1 - 22.
(d) For what value of x does f 1x2 = - 3?
(e) Solve f 1x2 7 0.
(f) Graph y = f 1x - 32.
1
(g) Graph y = f a xb.
2
(h) Graph y = - f 1x2.
14. f 1x2 =
f 1x + h2 - f 1x2
h
x + 1
1
; g1x2 =
x - 1
x
, h ≠ 0.
y
4
5
(2, 1)
(4, 3)
(3, 3)
(0, 0)
4
5
x
Chapter Review
17. Use the graph of the function f shown to find:
(a) The domain and the range of f .
(b) The intervals on which f is increasing, decreasing, or
constant.
(c) The local minimum values and local maximum values.
(d) The absolute maximum and absolute minimum.
(e) Whether the graph is symmetric with respect to the
x-axis, the y-axis, or the origin.
(f) Whether the function is even, odd, or neither.
(g) The intercepts, if any.
y
4
(3, 0)
(22, 1)
26
(24,23) (23, 0)
269
(4, 3)
6 x
(2, 21)
24
In Problems 18–21, determine (algebraically) whether the given function is even, odd, or neither.
18. f 1x2 = x3 - 4x
19. g1x2 =
4 + x2
1 + x4
21. f 1x2 =
20. G1x2 = 1 - x + x3
x
1 + x2
In Problems 22 and 23, use a graphing utility to graph each function over the indicated interval. Approximate any local maximum values
and local minimum values. Determine where the function is increasing and where it is decreasing.
22. f 1x2 = 2x3 - 5x + 1
1 - 3, 32
24. Find the average rate of change of f 1x2 = 8x - x:
(a) From 1 to 2
(b) From 0 to 1
(c) From 2 to 4
23. f 1x2 = 2x4 - 5x3 + 2x + 1
1 - 2, 32
2
In Problems 25 and 26, find the average rate of change from 2 to 3 for each function f. Be sure to simplify.
25. f 1x2 = 2 - 5x
26. f 1x2 = 3x - 4x2
In Problems 27 and 28, is the graph shown the graph of a function?
27.
28.
y
y
x
x
In Problems 29 and 30, graph each function. Be sure to label at least three points.
29. f 1x2 = 0 x 0
30. f 1x2 = 1x
In Problems 31–36, graph each function using the techniques of shifting, compressing or stretching, and reflections. Identify any
intercepts of the graph. State the domain and, based on the graph, find the range.
31. F 1x2 = 0 x 0 - 4
32. g1x2 = - 2 0 x 0
33. h1x2 = 2x - 1
34. f 1x2 = 21 - x
35. h1x2 = 1x - 12 2 + 2
36. g1x2 = - 21x + 22 3 - 8
In Problems 37 and 38:
(a) Find the domain of each function.
(d) Based on the graph, find the range.
37. f1x2 = b
3x
x + 1
(b) Locate any intercepts.
(e) Is f continuous on its domain?
if - 2 6 x … 1
if x 7 1
39. A function f is defined by
f 1x2 =
x
38. f 1x2 = c 1
3x
(c) Graph each function.
if - 4 … x 6 0
if x = 0
if x 7 0
Ax + 5
6x - 2
If f 112 = 4, find A.
40. Constructing a Closed Box A closed box with a square base
is required to have a volume of 10 cubic feet.
(a) Build a model that expresses the amount A of material
used to make such a box as a function of the length x of
a side of the square base.
(b) How much material is required for a base 1 foot by
1 foot?
(c) How much material is required for a base 2 feet by
2 feet?
(d) Graph A = A1x2. For what value of x is A smallest?
41. Area of a Rectangle A rectangle has one vertex in quadrant I
on the graph of y = 10 - x2, another at the origin, one on
the positive x-axis, and one on the positive y-axis.
(a) Express the area A of the rectangle as a function of x.
(b) Find the largest area A that can be enclosed by the
rectangle.
270
CHAPTER 3 Functions and Their Graphs
The Chapter Test Prep Videos are step-by-step solutions available in
Channel. Flip back to the Resources
, or on this text’s
for Success page for a link to this text’s YouTube channel.
Chapter Test
1. Determine whether each relation represents a function. For
each function, state the domain and the range.
(a) 5 12, 52, 14, 62, 16, 72, 18, 82 6
(b) 5 11, 32, 14, - 22, 1 - 3, 52, 11, 72 6
(c)
7. Consider the function g1x2 = b
y
6
(a)
(b)
(c)
(d)
4
2
x
4
2
2
4
4
y
4
2
x
2
2
4
2
3. g1x2 =
x + 2
0x + 20
x - 4
x2 + 5x - 36
5. Consider the graph of the function f below.
4. h1x2 =
4 (1, 3)
(0, 2)
(2, 0)
4
(a)
(b)
(c)
(d)
(e)
x
4
2
(5, 3)
4
8. For the function f 1x2 = 3x2 - 2x + 4, find the average
rate of change of f from 3 to 4.
11. The variable interest rate on a student loan changes each
July 1 based on the bank prime loan rate. For the years
1992–2007, this rate can be approximated by the model
(5, 2)
(3, 3)
Find the domain and the range of f.
List the intercepts.
Find f 112.
For what value(s) of x does f 1x2 = - 3?
Solve f 1x2 6 0.
where x is the number of years since 1992 and r is the
interest rate as a percent.
(a) Use a graphing utility to estimate the highest rate during
this time period. During which year was the interest rate
the highest?
(b) Use the model to estimate the rate in 2010. Does this
value seem reasonable?
Source: U.S. Federal Reserve
12. A community skating rink is in the shape of a rectangle with
semicircles attached at the ends. The length of the rectangle
is 20 feet less than twice the width. The thickness of the ice is
0.75 inch.
(a) Build a model that expresses the ice volume, V, as a
function of the width, x.
(b) How much ice is in the rink if the width is 90 feet?
y
(2, 0)
Graph the function.
List the intercepts.
Find g1 - 52.
Find g122.
r 1x2 = - 0.115x2 + 1.183x + 5.623,
In Problems 2–4, find the domain of each function and evaluate
each function at x = - 1.
2. f 1x2 = 24 - 5x
if x 6 - 1
if x Ú - 1
10. Graph each function using the techniques of shifting,
compressing or stretching, and reflecting. Start with the
graph of the basic function and show all stages.
(a) h1x2 = - 21x + 12 3 + 3
(b) g1x2 = 0 x + 4 0 + 2
6
4
2x + 1
x - 4
9. For the functions f 1x2 = 2x2 + 1 and g1x2 = 3x - 2, find
the following and simplify.
(a) 1f - g2 1x2
(b) 1f # g2 1x2
(c) f 1x + h2 - f 1x2
2
(d)
6. Use a graphing utility to graph the function
f 1x2 = - x4 + 2x3 + 4x2 - 2 on the interval 1 - 5, 52.
Then approximate any local maximum values and local
minimum values rounded to two decimal places. Determine
where the function is increasing and where it is decreasing.
Chapter Projects
271
Cumulative Review
In Problems 1–6, find the real solutions of each equation.
In Problems 11–14, graph each equation.
11. 3x - 2y = 12
12. x = y2
1. 3x - 8 = 10
2. 3x - x = 0
3. x - 8x - 9 = 0
4. 6x - 5x + 1 = 0
5. 0 2x + 3 0 = 4
15. For the equation 3x2 - 4y = 12, find the intercepts and
check for symmetry.
6. 22x + 3 = 2
16. Find the slope–intercept form of the equation of the line
containing the points 1 - 2, 42 and 16, 82 .
2
2
2
13. x2 + 1y - 32 2 = 16
14. y = 2x
In Problems 7–9, solve each inequality. Graph the solution set.
7. 2 - 3x 7 6
8. 0 2x - 5 0 6 3
9. 0 4x + 1 0 Ú 7
10. (a) Find the distance from P1 = 1 - 2, - 32 to P2 = 13, - 52.
(b) What is the midpoint of the line segment from P1 to P2?
(c) What is the slope of the line containing the points P1
and P2?
In Problems 17–19, graph each function.
17. f 1x2 = 1x + 22 2 - 3
18. f 1x2 =
1
x
19. f 1x2 = e
2 - x
0x0
if x … 2
if x 7 2
Chapter Projects
that include unlimited talk and text. The monthly cost is
primarily determined by the amount of data used and the
number of devices.
1. Suppose you expect to use 10 gigabytes of data for a
single smartphone. What would be the monthly cost of
each plan you are considering?
2. Suppose you expect to use 30 gigabytes of data and
want a personal hotspot, but you still have only a single
smartphone. What would be the monthly cost of each
plan you are considering?
3. Suppose you expect to use 20 gigabytes of data with
three smartphones sharing the data. What would be the
monthly cost of each plan you are considering?
4. Suppose you expect to use 20 gigabytes of data with a
single smartphone and a personal hotspot. What would
be the monthly cost of each plan you are considering?
5. Build a model that describes the monthly cost C, in
dollars, as a function of the number of data gigabytes
used, g, assuming a single smartphone and a personal
hotspot for each plan you are considering.
6. Graph each function from Problem 5.
Internet-based Project
I. Choosing a Wireless Data Plan Collect information from
your family, friends, or consumer agencies such as Consumer
Reports. Then decide on a cellular provider, choosing the
company that you feel offers the best service. Once you have
selected a service provider, research the various types of
individual plans offered by the company by visiting the
provider’s website. Many providers offer family plans
7.
Based on your particular usage, which plan is best for
you?
8. Now, develop an Excel spreadsheet to analyze the
various plans you are considering. Suppose you want
a family plan with unlimited talk and text that offers
10 gigabytes of shared data and costs $100 per month.
Additional gigabytes of data cost $15 per gigabyte, extra
phones can be added to the plan for $15 each per month,
and each hotspot costs $20 per month. Because wireless
272
CHAPTER 3 Functions and Their Graphs
data plans have a cost structure based on piecewise-defined functions, we need an “if/then” statement within Excel to analyze
the cost of the plan. Use the accompanying Excel spreadsheet as a guide in developing your spreadsheet. Enter into your
spreadsheet a variety of possible amounts of data and various numbers of additional phones and hotspots.
A
1
2 Monthly fee
3 Allotted data per month (GB)
4 Data used (GB)
5
6
7
8
9
10
B
D
$100
10
Cost per additional GB of data
12
$15
Monthly cost of hotspot
Number of hotspots
Monthly cost of additional phone
Number of additional phones
$20
1
$15
2
11
12 Cost of data
13 Cost of additional devices/hotspots
14
15 Total Cost
16
C
=IF(B4<B3,B2,B2+B5*(B4-B3))
=B8*B7+B10*B9
=B12+B13
9. Write a paragraph supporting the choice in plans that best meets your needs.
10. How are “if/then” loops similar to a piecewise-defined function?
Citation: Excel © 2013 Microsoft Corporation. Used with permission from Microsoft.
The following projects are available on the Instructor’s Resource Center (IRC).
II. Project at Motorola: Wireless Internet Service Use functions and their graphs to analyze the total cost of various wireless
Internet service plans.
III. Cost of Cable When government regulations and customer preference influence the path of a new cable line, the Pythagorean
Theorem can be used to assess the cost of installation.
IV. Oil Spill Functions are used to analyze the size and spread of an oil spill from a leaking tanker.
4
Linear and Quadratic
Functions
The Beta of a Stock
Investing in the stock market can be rewarding and fun, but how does
one go about selecting which stocks to purchase? Financial investment
firms hire thousands of analysts who track individual stocks (equities)
and assess the value of the underlying company. One measure the
analysts consider is the beta of the stock. Beta measures the risk
of an individual company’s equity relative to that of a market
basket of stocks, such as the Standard & Poor’s 500. But how is
beta computed?
—See the Internet-based Chapter Project I—
A Look Back
Up to now, our discussion has focused on graphs of equations and functions. We
learned how to graph equations using the point-plotting method, intercepts,
and the tests for symmetry. In addition, we learned what a function is and how to
identify whether a relation represents a function. We also discussed properties of
functions, such as domain/range, increasing/decreasing, even/odd, and average
rate of change.
A Look Ahead
Going forward, we will look at classes of functions. This chapter focuses on linear
and quadratic functions, their properties, and their applications.
Outline
4.1
4.2
4.3
4.4
4.5
Properties of Linear Functions and
Linear Models
Building Linear Models from Data
Quadratic Functions and Their
Properties
Build Quadratic Models from Verbal
Descriptions and from Data
Inequalities Involving Quadratic
Functions
Chapter Review
Chapter Test
Cumulative Review
Chapter Projects
273
273
274
CHAPTER 4 Linear and Quadratic Functions
4.1 Properties of Linear Functions and Linear Models
PREPARING FOR THIS SECTION Before getting started, review the following:
r Functions (Section 3.1, pp. 199–208)
r The Graph of a Function (Section 3.2, pp. 214–217)
r Properties of Functions (Section 3.3, pp. 223–231)
r Lines (Section 2.3, pp. 167–175)
r Graphs of Equations in Two Variables; Intercepts;
Symmetry (Section 2.2, pp. 157–164)
r Linear Equations (Section 1.1, pp. 82–87)
Now Work the ‘Are You Prepared?’ problems on page 280.
OBJECTIVES 1 Graph Linear Functions (p. 274)
2 Use Average Rate of Change to Identify Linear Functions (p. 274)
3 Determine Whether a Linear Function Is Increasing, Decreasing, or
Constant (p. 277)
4 Build Linear Models from Verbal Descriptions (p. 278)
1 Graph Linear Functions
In Section 2.3 we discussed lines. In particular, for nonvertical lines we
developed the slope–intercept form of the equation of a line y = mx + b. When the
slope–intercept form of a line is written using function notation, the result is a linear
function.
DEFINITION
A linear function is a function of the form
f1x2 = mx + b
The graph of a linear function is a line with slope m and y-intercept b. Its domain
is the set of all real numbers.
Functions that are not linear are said to be nonlinear.
EX AMPLE 1
Solution
y
(0, 7)
Δx 1
1
3
This is a linear function with slope m = - 3 and y-intercept b = 7. To graph
this function, plot the point 10, 72, the y-intercept, and use the slope to find an
additional point by moving right 1 unit and down 3 units. See Figure 1. The domain
and the range of f are each the set of all real numbers.
Alternatively, an additional point could have been found by evaluating the
function at some x ≠ 0. For x = 1, f 112 = - 3112 + 7 = 4 and the point 11, 42
lies on the graph.
(1, 4)
3
1
Graph the linear function f 1x2 = - 3x + 7. What are the domain and the range of f ?
r
Δy 3
5
Graphing a Linear Function
5 x
Figure 1 f 1x2 = - 3x + 7
Now Work
PROBLEMS
13(a)
AND
(b)
2 Use Average Rate of Change to Identify Linear Functions
Look at Table 1, which shows certain values of the independent variable x and
corresponding values of the dependent variable y for the function f 1x2 = - 3x + 7.
Notice that as the value of the independent variable, x, increases by 1, the value of
the dependent variable y decreases by 3. That is, the average rate of change of y with
respect to x is a constant, - 3.
SECTION 4.1 Properties of Linear Functions and Linear Models
Table 1
x
y = f(x) = −3x + 7
-2
13
Average Rate of Change =
275
𝚫y
𝚫x
10 - 13
-3
=
= -3
- 1 - 1 - 22
1
-1
10
-3
7 - 10
=
= -3
0 - 1 - 12
1
0
7
-3
1
4
2
1
3
-2
-3
-3
It is not a coincidence that the average rate of change of the linear function
∆y
f1x2 = - 3x + 7 is the slope of the linear function. That is,
= m = - 3. The
∆x
following theorem states this fact.
THEOREM
Average Rate of Change of a Linear Function
Linear functions have a constant average rate of change. That is, the average
rate of change of a linear function f 1x2 = mx + b is
∆y
= m
∆x
Proof The average rate of change of f1x2 = mx + b from x1 to x2, x1 ≠ x2, is
f1x2 2 - f1x1 2
1mx2 + b2 - 1mx1 + b2
∆y
=
=
x2 - x1
x2 - x1
∆x
=
m1x2 - x1 2
mx2 - mx1
=
= m
x2 - x1
x2 - x1
■
Based on the theorem just proved, the average rate of change of the function
2
2
g1x2 = - x + 5 is - .
5
5
Now Work
PROBLEM
13(C)
As it turns out, only linear functions have a constant average rate of change.
Because of this, the average rate of change can be used to determine whether a
function is linear. This is especially useful if the function is defined by a data set.
EXAMPL E 2
Using the Average Rate of Change to Identify Linear Functions
(a) A strain of E. coli known as Beu 397-recA441 is placed into a Petri dish at
30° Celsius and allowed to grow. The data shown in Table 2 on the next page are
collected. The population is measured in grams and the time in hours. Plot the
ordered pairs 1x, y2 in the Cartesian plane, and use the average rate of change
to determine whether the function is linear.
276
CHAPTER 4 Linear and Quadratic Functions
(b) The data in Table 3 represent the maximum number of heartbeats that a healthy
individual of different ages should have during a 15-second interval of time
while exercising. Plot the ordered pairs 1x, y2 in the Cartesian plane, and use
the average rate of change to determine whether the function is linear.
Table 3
Table 2
Maximum Number
of Heartbeats, y
Time
(hours), x
Population
(grams), y
(x, y)
Age, x
0
0.09
(0, 0.09)
20
50
(20, 50)
1
0.12
(1, 0.12)
30
47.5
(30, 47.5)
2
0.16
(2, 0.16)
40
45
(40, 45)
3
0.22
(3, 0.22)
50
42.5
(50, 42.5)
4
0.29
(4, 0.29)
60
40
(60, 40)
5
0.39
(5, 0.39)
70
37.5
(70, 37.5)
(x, y )
Source: American Heart Association
Solution
Compute the average rate of change of each function. If the average rate of change
is constant, the function is linear. If the average rate of change is not constant, the
function is nonlinear.
(a) Figure 2 shows the points listed in Table 2 plotted in the Cartesian plane. Note
that it is impossible to draw a straight line that contains all the points. Table 4
displays the average rate of change of the population.
Population (grams), y
y
Table 4
0.4
Time (hours), x
Population (grams), y
0
0.09
𝚫y
𝚫x
0.12 - 0.09
= 0.03
1 - 0
0.3
0.2
1
0.12
2
0.16
0.04
0.1
0
1
2
3
4
5
x
0.06
Time (hours), x
Figure 2
Average Rate of Change =
3
0.22
0.07
4
0.29
0.10
5
0.39
Because the average rate of change is not constant, the function is not linear.
In fact, because the average rate of change is increasing as the value of the
independent variable increases, the function is increasing at an increasing rate.
So not only is the population increasing over time, but it is also growing more
rapidly as time passes.
(b) Figure 3 shows the points listed in Table 3 plotted in the Cartesian plane. Note
that the data in Figure 3 lie on a straight line. Table 5 displays the average
rate of change of the maximum number of heartbeats. The average rate of
change of the heartbeat data is constant, - 0.25 beat per year, so the function is
linear.
SECTION 4.1 Properties of Linear Functions and Linear Models
Table 5
y
Heartbeats
50
Age, x
Maximum Number
of Heartbeats, y
20
50
Average Rate of Change =
277
𝚫y
𝚫x
47.5 - 50
= - 0.25
30 - 20
45
30
47.5
40
40
45
- 0.25
- 0.25
20
30
40
50
60
70
x
50
42.5
Age
- 0.25
60
Figure 3
40
- 0.25
70
37.5
r
Now Work
PROBLEM
21
3 Determine Whether a Linear Function Is Increasing,
Decreasing, or Constant
Look back at the Seeing the Concept on page 169. When the slope m of a linear
function is positive (m 7 0), the line slants upward from left to right. When the
slope m of a linear function is negative (m 6 0), the line slants downward from left
to right. When the slope m of a linear function is zero (m = 0), the line is horizontal.
THEOREM
Increasing, Decreasing, and Constant Linear Functions
A linear function f1x2 = mx + b is increasing over its domain if its slope, m,
is positive. It is decreasing over its domain if its slope, m, is negative. It is constant
over its domain if its slope, m, is zero.
EXAMPL E 3
Determining Whether a Linear Function Is Increasing,
Decreasing, or Constant
Determine whether the following linear functions are increasing, decreasing, or
constant.
(a) f1x2 = 5x - 2
3
(c) s 1t2 = t - 4
4
Solution
(b) g1x2 = - 2x + 8
(d) h 1z2 = 7
(a) For the linear function f1x2 = 5x - 2, the slope is 5, which is positive. The
function f is increasing on the interval 1 - q , q 2.
(b) For the linear function g1x2 = - 2x + 8, the slope is - 2, which is negative. The
function g is decreasing on the interval 1 - q , q 2.
3
3
(c) For the linear function s 1t2 = t - 4, the slope is , which is positive. The
4
4
function s is increasing on the interval 1 - q , q 2.
(d) The linear function h can be written as h 1z2 = 0z + 7. Because the slope is 0,
the function h is constant on the interval 1 - q , q 2.
r
Now Work
PROBLEM
13 (d)
278
CHAPTER 4 Linear and Quadratic Functions
4 Build Linear Models from Verbal Descriptions
When the average rate of change of a function is constant, a linear function can
model the relation between the two variables. For example, if a recycling company
pays $0.52 per pound for aluminum cans, then the relation between the price paid
p and the pounds recycled x can be modeled as the linear function p1x2 = 0.52x,
0.52 dollar
with slope m =
.
1 pound
Modeling with a Linear Function
If the average rate of change of a function is a constant m, a linear function f
can be used to model the relation between the two variables as follows:
f1x2 = mx + b
where b is the value of f at 0; that is, b = f102.
EX AMPLE 4
Straight-line Depreciation
Book value is the value of an asset that a company uses to create its balance sheet.
Some companies depreciate assets using straight-line depreciation so that the value
of the asset declines by a fixed amount each year. The amount of the decline depends
on the useful life that the company assigns to the asset. Suppose a company just
purchased a fleet of new cars for its sales force at a cost of $31,500 per car. The company
chooses to depreciate each vehicle using the straight-line method over 7 years. This
+31,500
means that each car will depreciate by
= +4500 per year.
7
(a) Write a linear function that expresses the book value V of each car as a function
of its age, x, in years.
(b) Graph the linear function.
(c) What is the book value of each car after 3 years?
(d) Interpret the slope.
(e) When will the book value of each car be $9000?
[Hint: Solve the equation V 1x2 = 9000.]
Solution
v
31,500
V 1x2 = - 4500x + 31,500
27,000
Book value ($)
(a) If we let V 1x2 represent the value of each car after x years, then V 102
represents the original value of each car, so V 102 = +31,500. The y-intercept of
the linear function is $31,500. Because each car depreciates by $4500 per year,
the slope of the linear function is - 4500. The linear function that represents the
book value V of each car after x years is
22,500
(b) Figure 4 shows the graph of V.
(c) The book value of each car after 3 years is
18,000
13,500
V 132 = - 4500132 + 31,500
= +18,000
9000
4500
1
2 3 4 5 6 7 x
Age of vehicle (years)
Figure 4 V 1 x2 = - 4500x + 31, 500
(d) Since the slope of V 1x2 = - 4500x + 31,500 is - 4500, the average rate of
change of the book value is - $4500/year. So for each additional year that passes,
the book value of the car decreases by $4500.
SECTION 4.1 Properties of Linear Functions and Linear Models
279
(e) To find when the book value will be $9000, solve the equation
V 1x2 = 9000
- 4500x + 31,500 = 9000
- 4500x = - 22,500
- 22,500
x =
= 5
- 4500
Subtract 31,500 from each side.
Divide by - 4500.
The car will have a book value of $9000 when it is 5 years old.
Now Work
EXAMPL E 5
PROBLEM
r
45
Supply and Demand
The quantity supplied of a good is the amount of a product that a company is willing
to make available for sale at a given price. The quantity demanded of a good is the
amount of a product that consumers are willing to purchase at a given price. Suppose
that the quantity supplied, S, and the quantity demanded, D, of cellular telephones
each month are given by the following functions:
S 1p2 = 60p - 900
D1p2 = - 15p + 2850
where p is the price (in dollars) of the telephone.
(a) The equilibrium price of a product is defined as the price at which quantity
supplied equals quantity demanded. That is, the equilibrium price is the price
at which S 1p2 = D1p2. Find the equilibrium price of cellular telephones.
What is the equilibrium quantity, the amount demanded (or supplied) at the
equilibrium price?
(b) Determine the prices for which quantity supplied is greater than quantity
demanded. That is, solve the inequality S 1p2 7 D1p2.
(c) Graph S = S 1p2 and D = D1p2, and label the equilibrium point, the point of
intersection of S and D.
Solution
(a) To find the equilibrium price, solve the equation S 1p2 = D1p2.
60p - 900 = - 15p + 2850
60p = - 15p + 3750
75p = 3750
p = 50
S1p2 = 60p - 900;
D (p) = - 15p + 2850
Add 900 to each side.
Add 15p to each side.
Divide each side by 75.
The equilibrium price is $50 per cellular phone. To find the equilibrium quantity,
evaluate either S 1p2 or D 1p2 at p = 50.
S 1502 = 601502 - 900 = 2100
The equilibrium quantity is 2100 cellular phones. At a price of $50 per phone,
the company will produce and sell 2100 phones each month and have no shortages
or excess inventory.
(b) The inequality S 1p2 7 D1p2 is
60p - 900
60p
75p
p
7
7
7
7
- 15p + 2850
- 15p + 3750
3750
50
S1p2 7 D1p2
Add 900 to each side.
Add 15p to each side.
Divide each side by 75.
If the company charges more than $50 per phone, quantity supplied will
exceed quantity demanded. In this case the company will have excess phones in
inventory.
280
CHAPTER 4 Linear and Quadratic Functions
(c) Figure 5 shows the graphs of S = S 1p2 and D = D1p2 with the equilibrium
point labeled.
Quantity supplied,
Quantity demanded
S, D
S 5 S(p)
3000 (0, 2850)
Equilibrium point
(50, 2100)
2000
D 5 D(p)
1000
(15, 0)
50
100 p
Price ($)
r
Figure 5 Supply and demand functions
Now Work
PROBLEM
39
4.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Graph y = x2 - 1. (pp. 157–164)
4. Solve: 60x - 900 = - 15x + 2850. (pp. 82–87)
2. Find the slope of the line joining the points 12, 52 and
5. If f 1x2 = x2 - 4, find f 1 - 22. (pp. 199–208)
6. True or False The graph of the function f 1x2 = x2 is
increasing on the interval 10, q 2. (pp. 214–217)
1 - 1, 32. (pp. 167–175)
3. Find the average rate of change of f 1x2 = 3x2 - 2, from
2 to 4. (pp. 223–231)
Concepts and Vocabulary
7. For the graph of the linear function f 1x2 = mx + b, m is
the
and b is the
.
8. If the slope m of the graph of a linear function is
the function is increasing over its domain.
11. What is the only type of function that has a constant average
rate of change?
(a) linear function
(b) quadratic function
(c) step function
(d) absolute value function
,
12. A car has 12,500 miles on its odometer. Say the car is driven an
average of 40 miles per day. Choose the model that expresses
the number of miles N that will be on its odometer after x days.
(a) N(x) = - 40x + 12,500 (b) N(x) = 40x - 12,500
(c) N(x) = 12,500x + 40
(d) N(x) = 40x + 12,500
9. True or False The slope of a nonvertical line is the average
rate of change of the linear function.
10. True or False The average rate of change of f 1x2 = 2x + 8
is 8.
Skill Building
In Problems 13–20, a linear function is given.
(a) Determine the slope and y-intercept of each function.
(b) Use the slope and y-intercept to graph the linear function.
(c) Determine the average rate of change of each function.
(d) Determine whether the linear function is increasing, decreasing, or constant.
13. f 1x2 = 2x + 3
14. g1x2 = 5x - 4
15. h1x2 = - 3x + 4
16. p1x2 = - x + 6
1
17. f 1x2 = x - 3
4
2
18. h1x2 = - x + 4
3
19. F 1x2 = 4
20. G1x2 = - 2
In Problems 21–28, determine whether the given function is linear or nonlinear. If it is linear, determine the slope.
21.
x
y = f (x)
-2
-1
22.
x
y = f (x)
4
-2
1
-1
23.
x
y = f (x)
1/4
-2
1/2
-1
24.
x
y = f (x)
-8
-2
-4
-3
-1
0
0
-2
0
1
0
0
0
4
1
-5
1
2
1
1
1
8
2
-8
2
4
2
0
2
12
SECTION 4.1 Properties of Linear Functions and Linear Models
25.
26.
x
y = f (x)
-2
- 26
-2
-1
-4
-1
x
27.
x
y = f (x)
-4
-2
- 3.5
-1
y = f (x)
28.
x
y = f (x)
8
-2
0
8
-1
1
0
2
0
-3
0
8
0
4
1
-2
1
- 2.5
1
8
1
9
2
- 10
2
-2
2
8
2
16
281
Applications and Extensions
29. Suppose that f 1x2 = 4x - 1 and g1x2 = - 2x + 5.
(a) Solve f 1x2 = 0.
(b) Solve f 1x2 7 0.
(c) Solve f 1x2 = g1x2. (d) Solve f 1x2 … g1x2.
(e) Graph y = f 1x2 and y = g1x2 and label the point that
represents the solution to the equation f 1x2 = g1x2.
30. Suppose that f 1x2 = 3x + 5 and g1x2 = - 2x + 15.
(a) Solve f 1x2 = 0.
(b) Solve f 1x2 6 0.
(c) Solve f 1x2 = g1x2. (d) Solve f 1x2 Ú g1x2.
(e) Graph y = f 1x2 and y = g1x2 and label the point that
represents the solution to the equation f 1x2 = g1x2.
31. In parts (a)–(f), use the following figure.
y
y f (x)
34. In parts (a) and (b), use the following figure.
y
y f (x)
(2, 5)
x
y g(x)
(a) Solve the equation: f 1x2 = g1x2.
(b) Solve the inequality: f 1x2 … g1x2.
35. In parts (a) and (b), use the following figure.
y f (x )
y
(88, 80)
(0, 12)
(40, 50)
y h (x )
x
(40, 0)
(a) Solve f 1x2 = 50.
(c) Solve f 1x2 = 0.
(e) Solve f 1x2 … 80.
(b) Solve f 1x2 = 80.
(d) Solve f 1x2 7 50.
(f) Solve 0 6 f 1x2 6 80.
32. In parts (a)–(f), use the following figure.
y g(x )
(5, 12)
y
x
(6,5)
(0,5)
y g (x )
(a) Solve the equation: f 1x2 = g1x2.
(b) Solve the inequality: g1x2 … f 1x2 6 h1x2.
36. In parts (a) and (b), use the following figure.
y
(15, 60)
(0, 7)
y h (x)
(4, 7)
(5, 20)
x
(15, 0)
x
(a) Solve g1x2 = 20.
(c) Solve g1x2 = 0.
(e) Solve g1x2 … 60.
(b) Solve g1x2 = 60.
(d) Solve g1x2 7 20.
(f) Solve 0 6 g1x2 6 60.
33. In parts (a) and (b), use the following figure.
y f(x)
y
y g (x )
(4, 6)
x
(a) Solve the equation: f 1x2 = g1x2.
(b) Solve the inequality: f 1x2 7 g1x2.
(0, 8)
(7,8)
y g (x)
y f (x)
(a) Solve the equation: f 1x2 = g1x2.
(b) Solve the inequality: g1x2 6 f 1x2 … h1x2.
37. Truck Rentals The cost C, in dollars, of a one-day truck rental
is modeled by the function C 1x2 = 0.35x + 45, where x is
the number of miles driven.
(a) What is the cost if you drive x = 40 miles?
(b) If the cost of renting the truck is $108, how many miles
did you drive?
(c) Suppose that you want the cost to be no more than $150.
What is the maximum number of miles that you can
drive?
(d) What is the implied domain of C?
282
CHAPTER 4 Linear and Quadratic Functions
38. Phone Charges The monthly cost C, in dollars, for calls
from the United States to Germany on a certain phone plan
is modeled by the function C 1x2 = 0.26x + 5, where x is
the number of minutes used.
(a) What is the cost if you talk on the phone for x = 50
minutes?
(b) Suppose that your monthly bill is $21.64. How many
minutes did you use the phone?
(c) Suppose that you budget yourself $50 per month for the
phone. What is the maximum number of minutes that
you can talk?
(d) What is the implied domain of C if there are 30 days in
the month?
39. Supply and Demand Suppose that the quantity supplied S
and the quantity demanded D of T-shirts at a concert are
given by the following functions:
S1p2 = - 600 + 50p
D1p2 = 1200 - 25p
where p is the price of a T-shirt.
(a) Find the equilibrium price for T-shirts at this concert.
What is the equilibrium quantity?
(b) Determine the prices for which quantity demanded is
greater than quantity supplied.
(c) What do you think will eventually happen to the price
of T-shirts if quantity demanded is greater than quantity
supplied?
40. Supply and Demand Suppose that the quantity supplied S
and the quantity demanded D of hot dogs at a baseball game
are given by the following functions:
S1p2 = - 2000 + 3000p
D1p2 = 10,000 - 1000p
where p is the price of a hot dog.
(a) Find the equilibrium price for hot dogs at the baseball
game. What is the equilibrium quantity?
(b) Determine the prices for which quantity demanded is
less than quantity supplied.
(c) What do you think will eventually happen to the price of hot
dogs if quantity demanded is less than quantity supplied?
41. Taxes The function T 1x2 = 0.151x - 90752 + 907.50
represents the tax bill T of a single person whose adjusted
gross income is x dollars for income between $9075 and
$36,900, inclusive, in 2014.
Source: Internal Revenue Service
(a) What is the domain of this linear function?
(b) What is a single filer’s tax bill if adjusted gross income is
$20,000?
(c) Which variable is independent and which is dependent?
(d) Graph the linear function over the domain specified in
part (a).
(e) What is a single filer’s adjusted gross income if the tax
bill is $3671.25?
42. Luxury Tax In 2011, major league baseball signed a labor
agreement with the players. In this agreement, any team whose
payroll exceeded $189 million in 2014 had to pay a luxury tax
of 50%. The linear function T 1p2 = 0.501p - 1892
describes the luxury tax T for a team whose payroll was p
(in millions of dollars).
Source: Major League Baseball
(a) What is the implied domain of this linear function?
(b) What was the luxury tax for the New York Yankees,
whose 2014 payroll was $203.4 million?
(c) Graph the linear function.
(d) What is the payroll of a team that pays a luxury tax of
$15.7 million?
The point at which a company’s profits equal zero is called the
company’s break-even point. For Problems 43 and 44, let R represent
a company’s revenue, let C represent the company’s costs, and let x
represent the number of units produced and sold each day.
(a) Find the firm’s break-even point; that is, find x so that R = C.
(b) Find the values of x such that R 1x2 7 C 1x2. This represents
the number of units that the company must sell to earn a profit.
43. R 1x2 = 8x
C 1x2 = 4.5x + 17,500
44. R 1x2 = 12x
C 1x2 = 10x + 15,000
45. Straight-line Depreciation Suppose that a company has just
purchased a new computer for $3000. The company chooses
to depreciate the computer using the straight-line method
over 3 years.
(a) Write a linear model that expresses the book value V of
the computer as a function of its age x.
(b) What is the implied domain of the function found in
part (a)?
(c) Graph the linear function.
(d) What is the book value of the computer after 2 years?
(e) When will the computer have a book value of $2000?
46. Straight-line Depreciation Suppose that a company has just
purchased a new machine for its manufacturing facility for
$120,000. The company chooses to depreciate the machine
using the straight-line method over 10 years.
(a) Write a linear model that expresses the book value V of
the machine as a function of its age x.
(b) What is the implied domain of the function found in
part (a)?
(c) Graph the linear function.
(d) What is the book value of the machine after 4 years?
(e) When will the machine have a book value of $72,000?
47. Cost Function The simplest cost function is the linear
cost function, C 1x2 = mx + b, where the y-intercept b
represents the fixed costs of operating a business and the
slope m represents the cost of each item produced. Suppose
that a small bicycle manufacturer has daily fixed costs of
$1800, and each bicycle costs $90 to manufacture.
(a) Write a linear model that expresses the cost C of
manufacturing x bicycles in a day.
(b) Graph the model.
(c) What is the cost of manufacturing 14 bicycles in a day?
(d) How many bicycles could be manufactured for $3780?
48. Cost Function Refer to Problem 47. Suppose that the
landlord of the building increases the bicycle manufacturer’s
rent by $100 per month.
(a) Assuming that the manufacturer is open for business
20 days per month, what are the new daily fixed costs?
(b) Write a linear model that expresses the cost C of
manufacturing x bicycles in a day with the higher rent.
(c) Graph the model.
(d) What is the cost of manufacturing 14 bicycles in a day?
(e) How many bicycles can be manufactured for $3780?
49. Truck Rentals A truck rental company rents a truck for one
day by charging $31.95 plus $0.89 per mile.
(a) Write a linear model that relates the cost C, in dollars, of
renting the truck to the number x of miles driven.
SECTION 4.1 Properties of Linear Functions and Linear Models
(b) What is the cost of renting the truck if the truck is driven
110 miles? 230 miles?
50. International Call Plan A cell phone company offers an
international plan by charging $30 for the first 80 minutes,
plus $0.50 for each minute over 80.
283
(a) Write a linear model that relates the cost C, in dollars, of
talking x minutes, assuming x Ú 80.
(b) What is the cost of talking 105 minutes?
120 minutes?
Mixed Practice
51. Building a Linear Model from Data How many songs
can an iPod hold? The following data represent the
memory m and the number of songs, n.
Memory, m (gigabytes)
Number of Songs, n
8
16
32
64
1750
3500
7000
14,000
(a) Plot the ordered pairs 1m, n2 in a Cartesian plane.
(b) Show that the number of songs n is a linear function of
the memory m.
(c) Determine the linear function that describes the
relation between m and n.
(d) What is the implied domain of the linear function?
(e) Graph the linear function in the Cartesian plane drawn
in part (a).
(f) Interpret the slope.
52. Building a Linear Model from Data The following data
represent the various combinations of soda and hot dogs
that Yolanda can buy at a baseball game with $60.
Soda, s
Hot Dogs, h
20
15
10
5
0
3
6
9
(a) Plot the ordered pairs 1s, h2 in a Cartesian plane.
(b) Show that the number of hot dogs purchased h is a
linear function of the number of sodas purchased s.
(c) Determine the linear function that describes the
relation between s and h.
(d) What is the implied domain of the linear function?
(e) Graph the linear function in the Cartesian plane drawn
in part (a).
(f) Interpret the slope.
(g) Interpret the values of the intercepts.
Explaining Concepts: Discussion and Writing
53. Which of the following functions might have the graph
shown? (More than one answer is possible.)
(a) f 1x2 = 2x - 7
y
(b) g1x2 = - 3x + 4
(c) H1x2 = 5
(d) F 1x2 = 3x + 4
1
x
(e) G1x2 = x + 2
2
54. Which of the following functions might have the graph
shown? (More than one answer is possible.)
(a) f 1x2 = 3x + 1
y
(b) g1x2 = - 2x + 3
(c) H1x2 = 3
(d) F 1x2 = - 4x - 1
x
2
(e) G1x2 = - x + 3
3
55. Under what circumstances is a linear function f 1x2 = mx + b odd? Can a linear function ever be even?
56. Explain how the graph of f 1x2 = mx + b can be used to solve mx + b 7 0.
Retain Your Knowledge
Problems 57–60 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
57. Graph x2 - 4x + y2 + 10y - 7 = 0.
58. If f(x) =
2x + B
and f(5) = 8, what is the value of B?
x - 3
59. Find the average rate of change of f(x) = 3x2 - 5x from
1 to 3.
x … 0
x2
60. Graph g(x) = e
2x + 1 x 7 0
‘Are You Prepared?’ Answers
1.
2.
y
22 21
21
1 2
x
2
3
3. 18
4. 5506
5. 0
6. True
284
CHAPTER 4 Linear and Quadratic Functions
4.2 Building Linear Models from Data
PREPARING FOR THIS SECTION Before getting started, review the following:
r Rectangular Coordinates (Section 2.1, pp. 150–151)
r Functions (Section 3.1, pp. 199–208)
r Lines (Section 2.3, pp. 167–175)
Now Work the ‘Are You Prepared?’ problems on page 287.
OBJECTIVES 1 Draw and Interpret Scatter Diagrams (p. 284)
2 Distinguish between Linear and Nonlinear Relations (p. 285)
3 Use a Graphing Utility to Find the Line of Best Fit (p. 286)
1 Draw and Interpret Scatter Diagrams
In Section 4.1, we built linear models from verbal descriptions. Linear models can
also be constructed by fitting a linear function to data. The first step is to plot the
ordered pairs using rectangular coordinates. The resulting graph is a scatter diagram.
EX AMPLE 1
Drawing and Interpreting a Scatter Diagram
In baseball, the on-base percentage for a team represents the percentage of time that
the players safely reach base. The data given in Table 6 represent the number of runs
scored y and the on-base percentage x for teams in the National League during the
2013 baseball season.
Table 6
Team
On-Base
Percentage, x
Runs
Scored, y
Arizona
32.3
685
(x, y)
(32.3, 685)
Atlanta
32.1
688
(32.1, 688)
Chicago Cubs
30.0
602
(30.0, 602)
Cincinnati
32.7
698
(32.7, 698)
Colorado
32.3
706
(32.3, 706)
LA Dodgers
32.6
649
(32.6, 649)
Miami
29.3
513
(29.3, 513)
Milwaukee
31.1
640
(31.1, 640)
NY Mets
30.6
619
(30.6, 619)
Philadelphia
30.6
610
(30.6, 610)
Pittsburgh
31.3
634
(31.3, 634)
San Diego
30.8
618
(30.8, 618)
San Francisco
32.0
629
(32.0, 629)
St. Louis
33.2
783
(33.2, 783)
Washington
31.3
656
(31.3, 656)
Source: espn.go.com
(a) Draw a scatter diagram of the data, treating on-base percentage as the
independent variable.
(b) Use a graphing utility to draw a scatter diagram.
(c) Describe what happens to runs scored as the on-base percentage increases.
Solution
(a) To draw a scatter diagram, plot the ordered pairs listed in Table 6, with the
on-base percentage as the x-coordinate and the runs scored as the y-coordinate.
See Figure 6(a). Notice that the points in the scatter diagram are not connected.
SECTION 4.2 Building Linear Models from Data
285
(b) Figure 6(b) shows a scatter diagram using a TI-84 Plus C graphing calculator.
(c) The scatter diagrams show that as the on-base percentage increases, the number
of runs scored also increases.
Runs Scored versus On-base Percentage
in the National League, 2013
y
850
Runs Scored
800
850
750
700
650
600
550
500
29
450
0
29
29.5
30
30.5 31 31.5 32
On-base Percentage
32.5
33
33.5 x
(a)
Figure 6
Now Work
PROBLEM
34
r
(b)
11(a)
2 Distinguish between Linear and Nonlinear Relations
Notice that the points in Figure 6 do not follow a perfect linear relation (as they
do in Figure 3 in Section 4.1). However, the data do exhibit a linear pattern. There
are numerous possible explanations why the data are not perfectly linear, but one
easy explanation is the fact that other variables besides on-base percentage (such as
number of home runs hit) play a role in determining runs scored.
Scatter diagrams are used to help us to see the type of relation that exists
between two variables. In this text, we will discuss a variety of different relations
that may exist between two variables. For now, we concentrate on distinguishing
between linear and nonlinear relations. See Figure 7.
Figure 7
(a) Linear
y mx b, m 0
EXAMPL E 2
(b) Linear
y mx b, m 0
(c) Nonlinear
(d) Nonlinear
(e) Nonlinear
Distinguishing between Linear and Nonlinear Relations
Determine whether the relation between the two variables in each scatter diagram
in Figure 8 is linear or nonlinear.
Figure 8
Solution
(a)
(a) Linear
(b)
(b) Nonlinear
Now Work
PROBLEM
(c)
(c) Nonlinear
5
(d)
(d) Nonlinear
r
CHAPTER 4 Linear and Quadratic Functions
This section considers data whose scatter diagrams suggest that a linear relation
exists between the two variables.
Suppose that the scatter diagram of a set of data appears to indicate a linear
relationship, as in Figure 7(a) or (b). We might want to model the data by finding
an equation of a line that relates the two variables. One way to obtain a model for
such data is to draw a line through two points on the scatter diagram and determine
the equation of the line.
EX A MPLE 3
Finding a Model for Linearly Related Data
Use the data in Table 6 from Example 1.
(a) Select two points and find an equation of the line containing the points.
(b) Graph the line on the scatter diagram obtained in Example 1(a).
Solution
(a) Select two points, say 130.6, 6102 and 132.1, 6882 . The slope of the line joining
the points 130.6, 6102 and 132.1, 6882 is
m =
78
688 - 610
=
= 52
32.1 - 30.6
1.5
The equation of the line that has slope 52 and passes through 130.6, 6102 is
found using the point–slope form with m = 52, x1 = 30.6, and y1 = 610.
y - y1 = m1x - x1 2
Point–slope form of a line
y - 610 = 521x - 30.62
x1 = 30.6, y1 = 610, m = 52
y - 610 = 52x - 1591.2
y = 52x - 981.2
The model
(b) Figure 9 shows the scatter diagram with the graph of the line found in part (a).
Runs Scored versus On-base Percentage
in the National League, 2013
y
850
800
Runs Scored
286
750
(32.1, 688)
700
650
600
(30.6, 610)
550
500
0
29
29.5
30
Figure 9
30.5 31 31.5 32
On-base Percentage
32.5
33
33.5 x
r
Select two other points and complete the solution. Graph the line on the
scatter diagram obtained in Figure 6.
Now Work
PROBLEMS
11(b)
AND
(c)
3 Use a Graphing Utility to Find the Line of Best Fit
The model obtained in Example 3 depends on the selection of points, which will
vary from person to person. So the model that we found might be different from
the model you found. Although the model in Example 3 appears to fit the data well,
SECTION 4.2 Building Linear Models from Data
287
there may be a model that “fits them better.” Do you think your model fits the data
better? Is there a line of best fit? As it turns out, there is a method for finding a
model that best fits linearly related data (called the line of best fit).*
EXAMPL E 4
Finding a Model for Linearly Related Data
Use the data in Table 6 from Example 1.
(a) Use a graphing utility to find the line of best fit that models the relation between
on-base percentage and runs scored.
(b) Graph the line of best fit on the scatter diagram obtained in Example 1(b).
(c) Interpret the slope.
(d) Use the line of best fit to predict the number of runs a team will score if their
on-base percentage is 33.1.
Solution
(a) Graphing utilities contain built-in programs that find the line of best fit for a
collection of points in a scatter diagram. Executing the LINear REGression
program provides the results shown in Figure 10. This output shows the equation
y = ax + b, where a is the slope of the line and b is the y-intercept. The line of
best fit that relates on-base percentage to runs scored may be expressed as the
line
y = 49.40x - 906.29
The model
(b) Figure 11 shows the graph of the line of best fit, along with the scatter diagram.
(c) The slope of the line of best fit is 49.40, which means that, for every 1 percent
increase in the on-base percentage, runs scored increase by 49.40, on average.
(d) Letting x = 33.1 in the equation of the line of best fit, we obtain
y = 49.40133.12 - 906.29 ≈ 729 runs.
Figure 10
r
850
29
450
Now Work
34
Figure 11
PROBLEMS
11(d)
AND
(e)
Does the line of best fit appear to be a good fit? In other words, does it appear
to accurately describe the relation between on-base percentage and runs scored?
And just how “good” is this line of best fit? Look again at Figure 10. The
last line of output is r = 0.896. This number, called the correlation coefficient, r,
- 1 … r … 1, is a measure of the strength of the linear relation that exists between
two variables. The closer r is to 1, the more nearly perfect the linear relationship
is. If r is close to 0, there is little or no linear relationship between the variables. A
negative value of r, r 6 0, indicates that as x increases, y decreases; a positive value
of r, r 7 0, indicates that as x increases, y does also. The data given in Table 6, which
have a correlation coefficient of 0.896, are indicative of a linear relationship with
positive slope.
4.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Plot the points (1, 5), (2, 6), (3, 9), (1, 12) in the Cartesian
plane. Is the relation 5 11, 52, 12, 62, 13, 92, 11, 122 6 a
function? Why? (pp. 150 and 199–208)
2. Find an equation of the line containing the points 11, 42 and
13, 82 . (pp. 167–175)
Concepts and Vocabulary
3. A
is used to help us to see what type of
relation, if any, may exist between two variables.
4. True or False The correlation coefficient is a measure of the
strength of a linear relation between two variables and must
lie between - 1 and 1, inclusive.
* We shall not discuss the underlying mathematics of lines of best fit in this text.
288
CHAPTER 4 Linear and Quadratic Functions
Skill Building
In Problems 5–10, examine the scatter diagram and determine whether the type of relation is linear or nonlinear.
5. y
6. y
35
30
25
20
15
10
5
14
12
10
8
6
4
2
7.
22
−2
0 2 4 6 8 10 12 14 16 x
0 5 10 1520 2530 3540 x
8.
9.
55
10.
25
5
20
35
0
0
35
12
0
0
0
10
45
*In Problems 11–16:
(a) Draw a scatter diagram.
(b) Select two points from the scatter diagram, and find the equation of the line containing the points selected.
(c) Graph the line found in part (b) on the scatter diagram.
(d) Use a graphing utility to find the line of best fit.
(e) Use a graphing utility to draw the scatter diagram and graph the line of best fit on it.
11.
14.
x
3 4
5
6
7
8
y
4 6
7
x
-2
-1
0
1
2
y
7
6
3
2
0
12.
9
10 12 14 16
15.
x
3 5
7
9 11
6
9
13.
13
y
0 2
3
11
x
- 20
- 17
- 15
- 14
- 10
y
100
120
118
130
140
16.
x
-2 -1
0
1
2
y
-4
0
1
4
5
x
- 30
- 27
- 25
- 20
- 14
y
10
12
13
13
18
Applications and Extensions
17. Candy The following data represent the weight (in grams)
of various candy bars and the corresponding number of calories.
Weight, x
Calories, y
Hershey’s Milk Chocolate®
44.28
230
Nestle’s Crunch®
44.84
230
Butterfinger®
61.30
270
Baby Ruth®
66.45
Almond Joy®
47.33
Candy Bar
(d) Graph the line on the scatter diagram drawn in part (a).
(e) Use the linear model to predict the number of calories
in a candy bar that weighs 62.3 grams.
(f) Interpret the slope of the line found in part (c).
18. Raisins The following data represent the weight (in grams)
of a box of raisins and the number of raisins in the box.
Weight (grams), w
Number of Raisins, N
280
42.3
87
220
42.7
91
93
Twix® (with caramel)
58.00
280
42.8
Snickers®
61.12
280
42.4
87
210
42.6
89
Source: Megan Pocius, student at Joliet Junior College
42.4
90
42.3
82
42.5
86
42.7
86
42.5
86
Heath®
39.52
(a) Draw a scatter diagram of the data, treating weight as
the independent variable.
(b) What type of relation appears to exist between the
weight of a candy bar and the number of calories?
(c) Select two points and find a linear model that contains
the points.
Source: Jennifer Maxwell, student at Joliet
Junior College
SECTION 4.2 Building Linear Models from Data
(a) Draw a scatter diagram of the data, treating weight as
the independent variable.
(b) What type of relation appears to exist between the
weight of a box of raisins and the number of raisins?
(c) Select two points and find a linear model that contains
the points.
(d) Graph the line on the scatter diagram drawn in part (b).
(e) Use the linear model to predict the number of raisins in
a box that weighs 42.5 grams.
(f) Interpret the slope of the line found in part (c).
(d) Predict the head circumference of a child who is 26 inches
tall.
(e) What is the height of a child whose head circumference
is 17.4 inches?
Height, H
(inches)
Head Circumference, C
(inches)
25.25
16.4
25.75
16.9
25
16.9
27.75
17.6
26.5
17.3
27
17.5
26.75
17.3
26.75
17.5
27.5
17.5
19. Video Games and Grade-Point Average Professor Grant
Alexander wanted to find a linear model that relates the
number of hours a student plays video games each week,
h, to the cumulative grade-point average, G, of the student.
He obtained a random sample of 10 full-time students at his
college and asked each student to disclose the number of
hours spent playing video games and the student’s cumulative
grade-point average.
Hours of Video Games
per Week, h
Grade-point
Average, G
0
3.49
0
3.05
2
3.24
3
2.82
3
3.19
5
2.78
8
2.31
8
2.54
10
2.03
12
2.51
Source: Denise Slucki, student at Joliet
Junior College
21. Flight Time and Ticket Price The following data represent
nonstop flight time (in minutes) and one-way ticket price
(in dollars) for flying from Chicago to various cities on
Southwest Airlines.
City
20. Height versus Head Circumference A pediatrician wanted
to find a linear model that relates a child’s height, H, to head
circumference, C. She randomly selects nine children from
her practice, measures their height and head circumference,
and obtains the data shown. Let H represent the independent
variable and C the dependent variable.
(a) Use a graphing utility to draw a scatter diagram.
(b) Use a graphing utility to find the line of best fit
that models the relation between height and head
circumference. Express the model using function notation.
(c) Interpret the slope.
Price
Time
(minutes), t (dollars), P
Atlanta, GA
110
140
Los Angeles, CA
260
199
80
108
125
130
Nashville, TN
(a) Explain why the number of hours spent playing video
games is the independent variable and cumulative
grade-point average is the dependent variable.
(b) Use a graphing utility to draw a scatter diagram.
(c) Use a graphing utility to find the line of best fit that
models the relation between number of hours of
video game playing each week and grade-point average.
Express the model using function notation.
(d) Interpret the slope.
(e) Predict the grade-point average of a student who plays
video games for 8 hours each week.
(f) How many hours of video game playing do you think a
student plays whose grade-point average is 2.40?
289
Oklahoma City, OK
Omaha, NE
85
89
Ft. Myers, FL
170
175
Phoenix, AZ
225
191
Houston, TX
155
147
Seattle, WA
265
196
65
93
St. Louis, MO
Source: Southwest.com, for midmorning flights in
May 2014
(a) Use a graphing utility to draw a scatter diagram.
(b) Use a graphing utility to find the line of best fit that
models the relation between flight time and airfare.
Express the model using function notation.
(c) Interpret the slope.
(d) Predict the airfare for a flight from Chicago to Kansas
City, Missouri, if the flight time is 85 minutes. Round to
the nearest dollar.
(e) Predict the flight time from Chicago to Baltimore,
Maryland if the airfare is $120. Round to the nearest
minute.
290
CHAPTER 4 Linear and Quadratic Functions
Explaining Concepts: Discussion and Writing
22. Maternal Age versus Down Syndrome A biologist would
like to know how the age of the mother affects the incidence
of Down syndrome. The data to the right represent the age
of the mother and the incidence of Down syndrome per
1000 pregnancies. Draw a scatter diagram treating age of the
mother as the independent variable. Would it make sense to
find the line of best fit for these data? Why or why not?
Age of Mother, x
Incidence of Down
Syndrome, y
33
2.4
34
3.1
23. Find the line of best fit for the ordered pairs 11, 52 and
13, 82 . What is the correlation coefficient for these data?
Why is this result reasonable?
35
4
36
5
37
6.7
24. What does a correlation coefficient of 0 imply?
38
8.3
25. Explain why it does not make sense to interpret the y-intercept
in Problem 17.
39
10
40
13.3
41
16.7
42
22.2
43
28.6
44
33.3
45
50
26. Refer to Problem 19. Solve G1h2 = 0. Provide an
interpretation of this result. Find G102 . Provide an
interpretation of this result.
Source: Hook, E.B., Journal of the American
Medical Association, 249, 2034-2038, 1983.
Retain Your Knowledge
Problems 27–30 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
27. Find an equation for the line containing the points (–1, 5)
and (3, –3). Express your answer using either the general
form or the slope-intercept form of the equation of a line,
whichever you prefer.
x - 1
28. Find the domain of f(x) = 2
.
x - 25
29. For f(x) = 5x - 8 and g(x) = x2 - 3x + 4, find (g - f)(x).
30. Write the function whose graph is the graph of y = x2, but shifted
to the left 3 units and shifted down 4 units.
‘Are You Prepared?’ Answers
1.
No, because the input, 1,
corresponds to two different
outputs.
y
12
9
2. y = 2x + 2
6
3
1
2
3 x
4.3 Quadratic Functions and Their Properties
PREPARING FOR THIS SECTION Before getting started, review the following:
r Intercepts (Section 2.2, pp. 159–160)
r Graphing Techniques: Transformations (Section 3.5, pp. 247–256)
r Completing the Square (Section R.5, p. 56)
r Quadratic Equations (Section 1.2, pp. 92–99)
Now Work the ‘Are You Prepared?’ problems on page 299.
OBJECTIVES 1
2
3
4
5
Graph a Quadratic Function Using Transformations (p. 292)
Identify the Vertex and Axis of Symmetry of a Quadratic Function (p. 294)
Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts (p. 294)
Find a Quadratic Function Given Its Vertex and One Other Point (p. 297)
Find the Maximum or Minimum Value of a Quadratic Function (p. 298)
SECTION 4.3 Quadratic Functions and Their Properties
291
Quadratic Functions
Here are some examples of quadratic functions.
F1x2 = 3x2 - 5x + 1
DEFINITION
g1x2 = - 6x2 + 1
H1x2 =
1 2
2
x + x
2
3
A quadratic function is a function of the form
f1x2 = ax2 + bx + c
where a, b, and c are real numbers and a ≠ 0. The domain of a quadratic
function is the set of all real numbers.
In Words
A quadratic function is a function
defined by a second-degree
polynomial in one variable.
Many applications require a knowledge of quadratic functions. For example,
suppose that Texas Instruments collects the data shown in Table 7, which relate
the number of calculators sold to the price p (in dollars) per calculator. Since the
price of a product determines the quantity that will be purchased, we treat price as
the independent variable. The relationship between the number x of calculators sold
and the price p per calculator is given by the linear equation
x = 21,000 - 150p
Price p per Calculator,
Number of
(in dollars)
Calculators, x
Table 7
60
12,000
65
11,250
70
10,500
75
9,750
80
9,000
85
8,250
90
7,500
Then the revenue R derived from selling x calculators at the price p per calculator
is equal to the unit selling price p of the calculator times the number x of units
actually sold. That is,
R = xp
R 1p2 = 121,000 - 150p2p x = 21,000 - 150p
= - 150p2 + 21,000p
So the revenue R is a quadratic function of the price p. Figure 12 illustrates the
graph of this revenue function, whose domain is 0 … p … 140, since both x and p
must be nonnegative.
R
800,000
Revenue (dollars)
700,000
600,000
500,000
400,000
300,000
200,000
100,000
Figure 12
R(p) = - 150p2 + 21,000p
0
14
28
42 56 70 84 98 112 126 140
Price per calculator (dollars)
p
292
CHAPTER 4 Linear and Quadratic Functions
A second situation in which a quadratic function appears involves the motion
of a projectile. Based on Newton’s Second Law of Motion (force equals mass times
acceleration, F = ma), it can be shown that, ignoring air resistance, the path of a
projectile propelled upward at an inclination to the horizontal is the graph of a
quadratic function. See Figure 13 for an illustration.
1 Graph a Quadratic Function Using Transformations
Figure 13
Path of a cannonball
We know how to graph the square function f1x2 = x2. Figure 14 shows the graph of
1
three functions of the form f1x2 = ax2, a 7 0, for a = 1, a = , and a = 3. Note
2
that the larger the value of a, the “narrower” the graph is, and the smaller the value
of a, the “wider” the graph is.
Figure 15 shows the graphs of f1x2 = ax2 for a 6 0. Notice that these graphs
are reflections about the x-axis of the graphs in Figure 14. Based on the results of
these two figures, general conclusions can be drawn about the graph of f1x2 = ax2.
First, as 0 a 0 increases, the graph is stretched vertically (becomes “taller”), and as
0 a 0 gets closer to zero, the graph is compressed vertically (becomes “shorter”).
Second, if a is positive, the graph opens “up,” and if a is negative, the graph
opens “down.”
y
y
4
f (x ) = 3x 2
4
f (x ) = x 2
f (x ) = 1–2 x 2
–4
–4
4
4
x
x
f (x ) = – 1–2 x 2
–4
f (x) = –x 2
f (x) = – 3x 2
–4
Figure 14
Axis of
symmetry
Vertex is
highest point
Vertex is
lowest point
Axis of
symmetry
D Opens up
E Opens down
a.0
a,0
Figure 16
Graphs of a quadratic function,
f 1x2 = ax2 + bx + c, a ≠ 0
EX AMPLE 1
Figure 15
The graphs in Figures 14 and 15 are typical of the graphs of all quadratic functions,
which are called parabolas.* Refer to Figure 16, where two parabolas are pictured.
The one on the left opens up and has a lowest point; the one on the right opens
down and has a highest point. The lowest or highest point of a parabola is called the
vertex. The vertical line passing through the vertex in each parabola in Figure 16 is
called the axis of symmetry (usually abbreviated to axis) of the parabola. Because
the parabola is symmetric about its axis, the axis of symmetry of a parabola can be
used to find additional points on the parabola.
The parabolas shown in Figure 16 are the graphs of a quadratic function
f1x2 = ax2 + bx + c, a ≠ 0. Notice that the coordinate axes are not included in
the figure. Depending on the values of a, b, and c, the axes could be placed anywhere.
The important fact is that the shape of the graph of a quadratic function will look
like one of the parabolas in Figure 16.
In the following example, techniques from Section 3.5 are used to graph a
quadratic function f 1x2 = ax2 + bx + c, a ≠ 0. The method of completing the
square is used to write the function f in the form f1x2 = a1x - h2 2 + k.
Graphing a Quadratic Function Using Transformations
Graph the function f1x2 = 2x2 + 8x + 5. Find the vertex and axis of symmetry.
* Parabolas will be studied using a geometric definition later in this text.
SECTION 4.3 Quadratic Functions and Their Properties
Solution
293
Begin by completing the square on the right side.
f1x2 = 2x2 + 8x + 5
= 21x2 + 4x2 + 5
Factor out the 2 from 2x 2 + 8x.
= 21x2 + 4x + 42 + 5 - 8
Complete the square of x 2 + 4x by
adding 4. Notice that the factor of 2
requires that 8 be added and subtracted.
= 21x + 22 2 - 3
The graph of f can be obtained from the graph of y = x2 in three stages, as shown
in Figure 17. Now compare this graph to the graph in Figure 16(a). The graph of
f1x2 = 2x2 + 8x + 5 is a parabola that opens up and has its vertex (lowest point)
at 1 - 2, - 32. Its axis of symmetry is the line x = - 2.
y
3
y
3
(⫺1, 1)
⫺2
(⫺1, 2)
(1, 1)
(0, 0) 2
x
(a) y ⫽ x 2
(1, 2)
⫺3
(0, 0)
⫺3
Figure 17
y
3
3
(⫺3, 2)
x
(⫺1, 2)
⫺3 (⫺2, 0)
3
x
Replace x
by x ⫹ 2;
Shift left
2 units
(b) y ⫽ 2 x 2
3
(⫺1, ⫺1)
(⫺3, ⫺1)
⫺3
⫺3
Multiply
by 2;
Vertical
stretch
Axis of y
Symmetry3
x ⫽ ⫺2
(c) y ⫽ 2 (x ⫹ 2) 2
Subtract 3;
Shift down
3 units
x
⫺3
Vertex
(⫺2, ⫺3)
(d) y ⫽ 2(x ⫹ 2)2 ⫺ 3
r
Now Work
PROBLEM
25
The method used in Example 1 can be used to graph any quadratic function
f1x2 = ax2 + bx + c, a ≠ 0, as follows:
f1x2 = ax2 + bx + c
= aax2 +
b
xb + c
a
Factor out a from ax 2 + bx.
= aax2 +
b
b2
b2
x +
b
+
c
aa
b
a
4a2
4a2
b2
Complete the square by adding 2 .
4a
Look closely at this step!
= aax +
b 2
b2
b + c 2a
4a
Factor ; simplify.
= aax +
b 2
4ac - b2
b +
2a
4a
c -
4a
b2
4ac - b 2
b2
= c#
=
4a
4a
4a
4a
These results lead to the following conclusion:
If h = -
4ac - b2
b
and k =
, then
2a
4a
f1x2 = ax2 + bx + c = a1x - h2 2 + k
(1)
The graph of f1x2 = a1x - h2 2 + k is the parabola y = ax2 shifted
horizontally h units (replace x by x - h) and vertically k units (add k). As a
result, the vertex is at 1h, k2 , and the graph opens up if a 7 0 and down if
a 6 0. The axis of symmetry is the vertical line x = h.
294
CHAPTER 4 Linear and Quadratic Functions
For example, compare equation (1) with the solution given in Example 1.
f 1x2 = 21x + 22 2 - 3
= 21x - 1 - 22 2 2 + 1 - 32
= a1x - h2 2 + k
Because a = 2, the graph opens up. Also, because h = - 2 and k = - 3, its vertex
is at 1 - 2, - 32.
2 Identify the Vertex and Axis of Symmetry
of a Quadratic Function
We do not need to complete the square to obtain the vertex. In almost every case,
it is easier to obtain the vertex of a quadratic function f by remembering that
b
its x-coordinate is h = - . The y-coordinate k can then be found by evaluating f
2a
b
b
at - . That is, k = f a - b .
2a
2a
Properties of the Graph of a Quadratic Function
f1x2 = ax2 + bx + c
Vertex = a -
b
b
, f a- b b
2a
2a
a ≠ 0
Axis of symmetry: the vertical line x = -
b
(2)
2a
Parabola opens up if a 7 0; the vertex is a minimum point.
Parabola opens down if a 6 0; the vertex is a maximum point.
EX AMPLE 2
Locating the Vertex without Graphing
Without graphing, locate the vertex and axis of symmetry of the parabola defined
by f1x2 = - 3x2 + 6x + 1. Does it open up or down?
Solution
For this quadratic function, a = - 3, b = 6, and c = 1. The x-coordinate of the
vertex is
h = -
b
6
= = 1
2a
21 - 32
The y-coordinate of the vertex is
k = f a-
b
b = f112 = - 3112 2 + 6112 + 1 = 4
2a
The vertex is located at the point 11, 42 . The axis of symmetry is the line x = 1.
Because a = - 3 6 0, the parabola opens down.
r
3 Graph a Quadratic Function Using Its Vertex, Axis,
and Intercepts
The location of the vertex and intercepts, along with knowledge of whether
the graph opens up or down, usually provides enough information to graph
f1x2 = ax2 + bx + c, a ≠ 0.
The y-intercept is the value of f at x = 0; that is, the y-intercept is f102 = c.
The x-intercepts, if there are any, are found by solving the quadratic equation
ax2 + bx + c = 0
SECTION 4.3 Quadratic Functions and Their Properties
295
This equation has two, one, or no real solutions, depending on whether the
discriminant b2 - 4ac is positive, 0, or negative. Depending on the value of the
discriminant, the graph of f has x-intercepts, as follows:
The x-Intercepts of a Quadratic Function
1. If the discriminant b2 - 4ac 7 0, the graph of f1x2 = ax2 + bx + c has
two distinct x-intercepts so it crosses the x-axis in two places.
2. If the discriminant b2 - 4ac = 0, the graph of f 1x2 = ax2 + bx + c has
one x-intercept so it touches the x-axis at its vertex.
3. If the discriminant b2 - 4ac 6 0, the graph of f1x2 = ax2 + bx + c has
no x-intercepts so it does not cross or touch the x-axis.
Figure 18 illustrates these possibilities for parabolas that open up.
Axis of symmetry
x=– b
2a
y
Axis of symmetry
x=– b
2a
y
x-intercept
x -intercept
– b ,f – b
2a
2a
(
Figure 18
f(x) = ax + bx + c, a 7 0
(
D
2
EXAMPL E 3
b2 –
x
Axis of symmetry
x=– b
2a
y
x
x -intercept
x
b
b
) (– 2a , f (– 2a ))
– b ,0
2a
(
))
4ac > 0
E
b 2 – 4ac = 0
One x-intercept
Two x-intercepts
F
b 2 – 4ac < 0
No x-intercepts
Graphing a Quadratic Function Using Its Vertex, Axis,
and Intercepts
(a) Use the information from Example 2 and the locations of the intercepts to graph
f1x2 = - 3x2 + 6x + 1.
(b) Determine the domain and the range of f.
(c) Determine where f is increasing and where it is decreasing.
Solution
(a) In Example 2, we found the vertex to be at 11, 42 and the axis of symmetry to
be x = 1. The y-intercept is found by letting x = 0. The y-intercept is f102 = 1.
The x-intercepts are found by solving the equation f1x2 = 0. This results in the
equation
- 3x2 + 6x + 1 = 0
a = - 3, b = 6, c = 1
The discriminant b - 4ac = 162 - 41 - 32 112 = 36 + 12 = 48 7 0, so the
equation has two real solutions and the graph has two x-intercepts. Use the
quadratic formula to find that
2
Axis of symmetry
x 1
y
(1, 4)
4
(0, 1)
4
(0.15, 0)
(2, 1)
4 x
(2.15, 0)
Figure 19 f(x) = - 3x2 + 6x + 1
x =
2
- b + 2b2 - 4ac
- 6 + 248
- 6 + 423
=
=
≈ - 0.15
2a
-6
-6
and
- 6 - 248
- 6 - 423
- b - 2b2 - 4ac
=
=
≈ 2.15
2a
-6
-6
The x-intercepts are approximately - 0.15 and 2.15.
The graph is illustrated in Figure 19. Notice how we used the y-intercept and
the axis of symmetry, x = 1, to obtain the additional point 12, 12 on the graph.
x =
296
CHAPTER 4 Linear and Quadratic Functions
(b) The domain of f is the set of all real numbers. Based on the graph, the range of f
is the interval 1 - q , 44.
(c) The function f is increasing on the interval 1 - q , 12 and decreasing on the
interval 11, q 2.
r
Graph the function in Example 3 by completing the square and using
transformations. Which method do you prefer?
Now Work
PROBLEM
33
If the graph of a quadratic function has only one x-intercept or no x-intercepts,
it is usually necessary to plot an additional point to obtain the graph.
EX AMPLE 4
Graphing a Quadratic Function Using Its Vertex, Axis,
and Intercepts
(a) Graph f1x2 = x2 - 6x + 9 by determining whether the graph opens up or
down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts,
if any.
(b) Determine the domain and the range of f.
(c) Determine where f is increasing and where it is decreasing.
Solution
(a) For f1x2 = x2 - 6x + 9, note that a = 1, b = - 6, and c = 9. Because
a = 1 7 0, the parabola opens up. The x-coordinate of the vertex is
h = -
Axis of symmetry
y
x3
b
-6
= = 3
2a
2112
The y-coordinate of the vertex is
(0, 9)
k = f 132 = 132 2 - 6132 + 9 = 0
(6, 9)
6
3
(3, 0)
6
x
Figure 20 f (x) = x2 - 6x + 9
The vertex is at 13, 02 . The axis of symmetry is the line x = 3. The y-intercept is
f102 = 9. Since the vertex 13, 02 lies on the x-axis, the graph touches the x-axis
at the x-intercept. By using the axis of symmetry and the y-intercept at (0, 9), we
can locate the additional point 16, 92 on the graph. See Figure 20.
(b) The domain of f is the set of all real numbers. Based on the graph, the range of f
is the interval 3 0, q 2.
(c) The function f is decreasing on the interval 1 - q , 32 and increasing on the
interval 13, q 2.
r
Now Work
EX AMPLE 5
PROBLEM
39
Graphing a Quadratic Function Using Its Vertex, Axis,
and Intercepts
(a) Graph f1x2 = 2x2 + x + 1 by determining whether the graph opens up or
down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts,
if any.
(b) Determine the domain and the range of f.
(c) Determine where f is increasing and where it is decreasing.
Solution
(a) For f1x2 = 2x2 + x + 1, we have a = 2, b = 1, and c = 1. Because a = 2 7 0,
the parabola opens up. The x-coordinate of the vertex is
h = -
b
1
= 2a
4
SECTION 4.3 Quadratic Functions and Their Properties
NOTE In Example 5, since the vertex is
above the x-axis and the parabola opens
up, we can conclude that the graph
of the quadratic function will have no
x-intercepts.
■
2
1
( – 1–2 , 1)
The y-coordinate of the vertex is
1
1
1
7
k = f a - b = 2a b + a - b + 1 =
4
16
4
8
1 7
1
The vertex is at a - , b . The axis of symmetry is the line x = - . The
4 8
4
y-intercept is f 102 = 1. The x-intercept(s), if any, satisfy the equation
2x2 + x + 1 = 0. The discriminant b2 - 4ac = 112 2 - 4122 112 = - 7 6 0.
This equation has no real solutions, which means the graph has no x-intercepts.
1
Use the point (0, 1) and the axis of symmetry x = - to locate the additional
4
1
point a - , 1b on the graph. See Figure 21.
2
y
Axis of
symmetry
x – 1–4
297
(b) The domain of f is the set of all real numbers. Based on the graph, the range of f
7
is the interval c , q b .
8
1
(c) The function f is decreasing on the interval a - q , - b and is increasing on the
4
1
interval a - , q b .
4
(0, 1)
( – 1–4 , 7–8 )
–1
x
1
r
Figure 21 f (x) = 2x + x + 1
2
Now Work
PROBLEM
43
4 Find a Quadratic Function Given Its Vertex and One Other Point
If the vertex 1h, k2 and one additional point on the graph of a quadratic
function f 1x2 = ax2 + bx + c, a ≠ 0, are known, then
f 1x2 = a1x - h2 2 + k
(3)
can be used to obtain the quadratic function.
EXAMPL E 6
Finding the Quadratic Function Given Its Vertex
and One Other Point
Determine the quadratic function whose vertex is 11, - 52 and whose y-intercept
is - 3.
Solution
The vertex is 11, - 52, so h = 1 and k = - 5. Substitute these values into equation (3).
y
h = 1, k = - 5
f1x2 = a1x - 12 2 - 5
4
(0, –3)
f1x2 = a1x - 12 - 5
To determine the value of a, use the fact that f102 = - 3 (the y-intercept).
8
–1
Equation (3)
2
12
–2
f1x2 = a1x - h2 2 + k
- 3 = a10 - 12 2 - 5
1
2
3
x = 0, y = f1 02 = - 3
-3 = a - 5
4 x
a = 2
–4
(1, –5)
–8
The quadratic function we seek is
f1x2 = a1x - h2 2 + k = 21x - 12 2 - 5 = 2x2 - 4x - 3
Figure 22 f 1x2 = 2x2 - 4x - 3
See Figure 22.
Now Work
PROBLEM
49
r
298
CHAPTER 4 Linear and Quadratic Functions
5 Find the Maximum or Minimum Value of a Quadratic Function
The graph of a quadratic function
f1x2 = ax2 + bx + c
a ≠ 0
b
b
, f a - b b . This vertex is the highest point on the
2a
2a
graph if a 6 0 and the lowest point on the graph if a 7 0. If the vertex is the highest
b
point 1a 6 02, then f a - b is the maximum value of f. If the vertex is the
2a
b
lowest point 1a 7 02, then f a - b is the minimum value of f.
2a
is a parabola with vertex at a -
EX AMPLE 7
Finding the Maximum or Minimum Value of a Quadratic Function
Determine whether the quadratic function
f1x2 = x2 - 4x - 5
has a maximum or a minimum value. Then find the maximum or minimum value.
Solution
Compare f 1x2 = x2 - 4x - 5 to f 1x2 = ax2 + bx + c. Then a = 1, b = - 4, and
c = - 5. Because a 7 0, the graph of f opens up, which means the vertex is a minimum
point. The minimum value occurs at
x = -
b
-4
4
= = = 2
2a
2112
2
c
a = 1, b = - 4
The minimum value is
f a-
b
b = f122 = 22 - 4122 - 5 = 4 - 8 - 5 = - 9
2a
Now Work
PROBLEM
57
SUMMARY
Steps for Graphing a Quadratic Function f(x) = ax2 + bx + c, a ≠ 0
Option 1
STEP 1: Complete the square in x to write the quadratic function in the form f1x2 = a1x - h2 2 + k.
STEP 2: Graph the function in stages using transformations.
Option 2
STEP 1: Determine whether the parabola opens up 1a 7 02 or down 1a 6 02.
b
b
STEP 2: Determine the vertex a - , f a - b b .
2a
2a
b
STEP 3: Determine the axis of symmetry, x = - .
2a
STEP 4: Determine the y-intercept, f 102 , and the x-intercepts, if any.
(a) If b2 - 4ac 7 0, the graph of the quadratic function has two x-intercepts, which are found by solving
the equation ax2 + bx + c = 0.
(b) If b2 - 4ac = 0, the vertex is the x-intercept.
(c) If b2 - 4ac 6 0, there are no x-intercepts.
STEP 5: Determine an additional point by using the y-intercept and the axis of symmetry.
STEP 6: Plot the points and draw the graph.
r
299
SECTION 4.3 Quadratic Functions and Their Properties
4.3 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. List the intercepts of the equation y = x2 - 9. (pp. 159–160)
3. To complete the square of x2 - 5x, you add the number
. (p. 56)
2. Find the real solutions of the equation 2x + 7x - 4 = 0.
(pp. 92–99)
2
4. To graph y = 1x - 42 2, you shift the graph of y = x2 to the
a distance of
units. (pp. 247–256)
Concepts and Vocabulary
5. The graph of a quadratic function is called a(n)
11. If b2 - 4ac 7 0, which of the following conclusions can be
made about the graph of f(x) = ax2 + bx + c, a ≠ 0?
(a) The graph has two distinct x-intercepts.
(b) The graph has no x-intercepts.
(c) The graph has three distinct x-intercepts.
(d) The graph has one x-intercept.
.
6. The vertical line passing through the vertex of a parabola is
called the
.
7. The x-coordinate of the vertex of f 1x2 = ax2 + bx + c,
a ≠ 0, is
.
8. True or False The graph of f 1x2 = 2x2 + 3x - 4 opens up.
12. If the graph of f(x) = ax2 + bx + c, a ≠ 0, has a maximum
value at its vertex, which of the following conditions must
be true?
b
b
(a) 7 0
(b) 6 0
2a
2a
(d) a 6 0
(c) a 7 0
9. True or False The y-coordinate of the vertex of
f 1x2 = - x2 + 4x + 5 is f 122 .
10. True or False If the discriminant b2 - 4ac = 0, the graph
of f 1x2 = ax2 + bx + c, a ≠ 0, will touch the x-axis at its
vertex.
Skill Building
In Problems 13–20, match each graph to one the following functions.
13. f 1x2 = x2 - 1
17. f 1x2 = x2 - 2x + 2
A.
14. f 1x2 = - x2 - 1
18. f 1x2 = x2 + 2x
B.
y
15. f 1x2 = x2 - 2x + 1
19. f 1x2 = x2 - 2x
C.
y
3
20. f 1x2 = x2 + 2x + 2
D.
y
2
2
16. f 1x2 = x2 + 2x + 1
y
3
(1, 1)
2
2 x
2
(1, 0)
E.
2 x
1
2
2 x
(0, 1)
3
2
(1, 1) 2
F.
y
1
2 x
G.
y
1
(1, 0)
3 x
1
2 x
1
y
2
2
(1, 1)
1
2
H.
y
3
3 x
2
(0, 1)
x
2
(1, 1)
In Problems 21–32, graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching,
and/or reflecting).
[Hint: If necessary, write f in the form f 1x2 = a1x - h2 2 + k.]
1
22. f 1x2 = 2x2 + 4
23. f 1x2 = 1x + 22 2 - 2
24. f 1x2 = 1x - 32 2 - 10
21. f 1x2 = x2
4
25. f 1x2 = x2 + 4x + 2
26. f 1x2 = x2 - 6x - 1
27. f 1x2 = 2x2 - 4x + 1
28. f 1x2 = 3x2 + 6x
1
2
4
29. f 1x2 = - x2 - 2x
30. f 1x2 = - 2x2 + 6x + 2
31. f 1x2 = x2 + x - 1
32. f 1x2 = x2 + x - 1
2
3
3
In Problems 33–48, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of
symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function
is increasing and where it is decreasing.
33. f 1x2 = x2 + 2x
37. f 1x2 = x + 2x - 8
2
34. f 1x2 = x2 - 4x
38. f 1x2 = x - 2x - 3
2
35. f 1x2 = - x2 - 6x
39. f 1x2 = x + 2x + 1
2
36. f 1x2 = - x2 + 4x
40. f 1x2 = x2 + 6x + 9
300
CHAPTER 4 Linear and Quadratic Functions
41. f 1x2 = 2x2 - x + 2
45. f 1x2 = 3x + 6x + 2
2
42. f 1x2 = 4x2 - 2x + 1
43. f 1x2 = - 2x2 + 2x - 3
46. f 1x2 = 2x + 5x + 3
47. f 1x2 = - 4x - 6x + 2
2
2
In Problems 49–54, determine the quadratic function whose graph is given.
49.
50.
y
y
2
–2
y
6
Vertex: (–3, 5)
4
4
2
–6
Vertex: (2, 1)
–3
1
2
(0, –1)
Vertex: (–1, –2)
51.
(0, 5)
1 x
–1
48. f 1x2 = 3x2 - 8x + 2
8
1
–3
44. f 1x2 = - 3x2 + 3x - 2
–2
x
(0, –4)
–1
1
2
3
4
5 x
–8
52.
53.
y
4
Vertex: (2, 3)
6
2
–1
1
(0, –1)
3
y
54.
y
8
Vertex: (–2, 6)
(3, 5)
4
4
5 x
6
2
2
–3
–4
x
3
–1
(–4, –2)
–1
1
x
–2
–4
–2
–4 Vertex: (1, –3)
In Problems 55–62, determine, without graphing, whether the given quadratic function has a maximum value or a minimum value, and
then find the value.
55. f 1x2 = 2x2 + 12x
59. f 1x2 = - x + 10x - 4
2
56. f1x2 = - 2x2 + 12x
60. f 1x2 = - 2x + 8x + 3
2
57. f 1x2 = 2x2 + 12x - 3
61. f 1x2 = - 3x + 12x + 1
2
58. f 1x2 = 4x2 - 8x + 3
62. f 1x2 = 4x2 - 4x
Mixed Practice
In Problems 63–70, (a) graph each function. (b) Determine the domain and the range of the function. (c) Determine where the function
is increasing and where it is decreasing.
2
3
63. f(x) = x2 - 2x - 15
64. g(x) = x2 - 2x - 8
65. h(x) = - x + 4
66. f (x) = x - 2
5
2
67. g(x) = - 2(x - 3)2 + 2
68. h(x) = - 3(x + 1)2 + 4
69. f(x) = 2x2 + x + 1
70. F(x) = - 4x2 + 20x - 25
Applications and Extensions
71. The graph of the function f 1x2 = ax2 + bx + c has vertex
at 10, 22 and passes through the point 11, 82. Find a, b, and c.
72. The graph of the function f 1x2 = ax2 + bx + c has vertex at
11, 42 and passes through the point 1 - 1, - 82. Find a, b, and c.
In Problems 73–78, for the given functions f and g:
(a) Graph f and g on the same Cartesian plane.
(b) Solve f 1x2 = g1x2.
(c) Use the result of part (b) to label the points of intersection of the graphs of f and g.
(d) Shade the region for which f 1x2 7 g1x2 —that is, the region below f and above g.
73. f 1x2 = 2x - 1; g1x2 = x2 - 4
75. f 1x2 = - x + 4; g1x2 = - 2x + 1
2
77. f 1x2 = - x2 + 5x; g1x2 = x2 + 3x - 4
74. f 1x2 = - 2x - 1; g1x2 = x2 - 9
76. f 1x2 = - x2 + 9; g1x2 = 2x + 1
78. f 1x2 = - x2 + 7x - 6; g1x2 = x2 + x - 6
Answer Problems 79 and 80 using the following: A quadratic function of the form f 1x2 = ax2 + bx + c with b2 - 4ac 7 0 may also be
written in the form f 1x2 = a1x - r1 2 1x - r2 2, where r1 and r2 are the x-intercepts of the graph of the quadratic function.
79. (a) Find a quadratic function whose x-intercepts are - 3
and 1 with a = 1; a = 2; a = - 2; a = 5.
(b) How does the value of a affect the intercepts?
(c) How does the value of a affect the axis of symmetry?
(d) How does the value of a affect the vertex?
(e) Compare the x-coordinate of the vertex with the
midpoint of the x-intercepts. What might you conclude?
80. (a) Find a quadratic function whose x-intercepts are - 5
and 3 with a = 1; a = 2; a = - 2; a = 5.
(b) How does the value of a affect the intercepts?
(c) How does the value of a affect the axis of symmetry?
(d) How does the value of a affect the vertex?
(e) Compare the x-coordinate of the vertex with the midpoint
of the x-intercepts. What might you conclude?
SECTION 4.3 Quadratic Functions and Their Properties
81. Suppose that f 1x2 = x2 + 4x - 21.
(a) What is the vertex of f?
(b) What are the x-intercepts of the graph of f ?
(c) Solve f 1x2 = - 21 for x. What points are on the graph
of f ?
(d) Use the information obtained in parts (a)–(c) to graph
f 1x2 = x2 + 4x - 21.
82. Suppose that f 1x2 = x2 + 2x - 8.
(a) What is the vertex of f ?
(b) What are the x-intercepts of the graph of f ?
(c) Solve f 1x2 = - 8 for x. What points are on the graph
of f ?
(d) Use the information obtained in parts (a)–(c) to graph
f 1x2 = x2 + 2x - 8.
83. Find the point on the line y = x that is closest to the
point 13, 12.
[Hint: Express the distance d from the point to the line as a
function of x, and then find the minimum value of 3d1x2 4 2.
84. Find the point on the line y = x + 1 that is closest to the
point 14, 12.
85. Maximizing Revenue Suppose that the manufacturer of
a gas clothes dryer has found that when the unit price is
p dollars, the revenue R (in dollars) is
R 1p2 = - 4p2 + 4000p
What unit price should be established for the dryer to
maximize revenue? What is the maximum revenue?
86. Maximizing Revenue The John Deere company has found
that the revenue, in dollars, from sales of riding mowers is a
function of the unit price p, in dollars, that it charges. If the
revenue R is
1
R 1p2 = - p2 + 1900p
2
what unit price p should be charged to maximize revenue?
What is the maximum revenue?
87. Minimizing Marginal Cost The marginal cost of a product
can be thought of as the cost of producing one additional
unit of output. For example, if the marginal cost of producing
the 50th product is $6.20, it cost $6.20 to increase production
from 49 to 50 units of output. Suppose the marginal cost C
(in dollars) to produce x thousand digital music players is
given by the function
C 1x2 = x2 - 140x + 7400
(a) How many players should be produced to minimize the
marginal cost?
(b) What is the minimum marginal cost?
88. Minimizing Marginal Cost (See Problem 87.) The marginal
cost C (in dollars) of manufacturing x cell phones (in thousands)
is given by
C 1x2 = 5x2 - 200x + 4000
301
(a) How many cell phones should be manufactured to
minimize the marginal cost?
(b) What is the minimum marginal cost?
89. Business The monthly revenue R achieved by selling
x wristwatches is figured to be R 1x2 = 75x - 0.2x2.
The monthly cost C of selling x wristwatches is
C 1x2 = 32x+ 1750
(a) How many wristwatches must the firm sell to maximize
revenue? What is the maximum revenue?
(b) Profit is given as P 1x2 = R 1x2 - C 1x2. What is the
profit function?
(c) How many wristwatches must the firm sell to maximize
profit? What is the maximum profit?
(d) Provide a reasonable explanation as to why the answers
found in parts (a) and (c) differ. Explain why a quadratic
function is a reasonable model for revenue.
90. Business The daily revenue R achieved by selling x boxes of
candy is figured to be R 1x2 = 9.5x - 0.04x2. The daily cost
C of selling x boxes of candy is C 1x2 = 1.25x + 250.
(a) How many boxes of candy must the firm sell to maximize
revenue? What is the maximum revenue?
(b) Profit is given as P 1x2 = R 1x2 - C 1x2. What is the
profit function?
(c) How many boxes of candy must the firm sell to maximize
profit? What is the maximum profit?
(d) Provide a reasonable explanation as to why the answers
found in parts (a) and (c) differ. Explain why a quadratic
function is a reasonable model for revenue.
91. Stopping Distance An accepted relationship between
stopping distance, d (in feet), and the speed of a car, v (in mph),
is d = 1.1v + 0.06v2 on dry, level concrete.
(a) How many feet will it take a car traveling 45 mph to stop
on dry, level concrete?
(b) If an accident occurs 200 feet ahead of you, what is the
maximum speed you can be traveling to avoid being
involved?
(c) What might the term 1.1v represent?
92. Birth Rate of Unmarried Women In the United States, the
birth rate B of unmarried women (births per 1000 unmarried
women) for women whose age is a is modeled by the function
B(a) = - 0.30a2 + 16.26a - 158.90
(a) What is the age of unmarried women with the highest
birth rate?
(b) What is the highest birth rate of unmarried women?
(c) Evaluate and interpret B 1402.
Source: National Vital Statistics System, 2013
93. Let f 1x2 = ax2 + bx + c, where a, b, and c are odd integers.
If x is an integer, show that f 1x2 must be an odd integer.
[Hint: x is either an even integer or an odd integer.]
Explaining Concepts: Discussion and Writing
94. Make up a quadratic function that opens down and has only
one x-intercept. Compare yours with others in the class.
What are the similarities? What are the differences?
97. State the circumstances that cause the graph of a quadratic
function f 1x2 = ax2 + bx + c to have no x-intercepts.
95. On one set of coordinate axes, graph the family of parabolas
f 1x2 = x2 + 2x + c for c = - 3, c = 0, and c = 1.
Describe the characteristics of a member of this family.
98. Why does the graph of a quadratic function open up if a 7 0
and down if a 6 0?
99. Can a quadratic function have a range of 1 - q , q 2? Justify
your answer.
96. On one set of coordinate axes, graph the family of parabolas
f 1x2 = x2 + bx + 1 for b = - 4, b = 0, and b = 4.
Describe the general characteristics of this family.
100. What are the possibilities for the number of times the graphs
of two different quadratic functions intersect?
302
CHAPTER 4 Linear and Quadratic Functions
Retain Your Knowledge
Problems 101–104 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
101. Determine whether x2 + 4y2 = 16 is symmetric respect to the x-axis, the y-axis, and/or the origin.
102. Solve the inequality 27 - x Ú 5x + 3. Write the solution in both set notation and interval notation.
103. Find the center and radius of the circle x2 + y2 - 10x + 4y + 20 = 0.
104. Write the function whose graph is the graph of y = 2x, but reflected about the y-axis.
‘Are You Prepared?’ Answers
1. 10, - 92, 1 - 3, 02, 13, 02
1
2. e - 4, f
2
3.
25
4
4. right; 4
4.4 Build Quadratic Models from Verbal
Descriptions and from Data
PREPARING FOR THIS SECTION Before getting started, review the following:
r Problem Solving (Section 1.7, pp. 134–140)
r Building Linear Models from Data
(Section 4.2, pp. 284–287)
Now Work the ‘Are You Prepared?’ problems on page 307.
OBJECTIVES 1 Build Quadratic Models from Verbal Descriptions (p. 302)
2 Build Quadratic Models from Data (p. 306)
In this section we will first discuss models in the form of a quadratic function when a
verbal description of the problem is given. We end the section by fitting a quadratic
function to data, which is another form of modeling.
When a mathematical model is in the form of a quadratic function, the
properties of the graph of the function can provide important information about the
model. In particular, we can use the quadratic function to determine the maximum
or minimum value of the function. The fact that the graph of a quadratic function
has a maximum or minimum value enables us to answer questions involving
optimization—that is, finding the maximum or minimum values in models.
1 Build Quadratic Models from Verbal Descriptions
In economics, revenue R, in dollars, is defined as the amount of money received
from the sale of an item and is equal to the unit selling price p, in dollars, of the item
times the number x of units actually sold. That is,
R = xp
The Law of Demand states that p and x are related: As one increases, the other
decreases. The equation that relates p and x is called the demand equation. When
the demand equation is linear, the revenue model is a quadratic function.
EX AMPLE 1
Maximizing Revenue
The marketing department at Texas Instruments has found that when certain
calculators are sold at a price of p dollars per unit, the number x of calculators sold
is given by the demand equation
x = 21,000 - 150p
(a) Find a model that expresses the revenue R as a function of the price p.
(b) What is the domain of R?
(c) What unit price should be used to maximize revenue?
(d) If this price is charged, what is the maximum revenue?
SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data
303
(e) How many units are sold at this price?
(f) Graph R.
(g) What price should Texas Instruments charge to collect at least $675,000 in revenue?
Solution
(a) The revenue is R = xp, where x = 21,000 - 150p.
R = xp = 121,000 - 150p2p = - 150p2 + 21,000p
The model
(b) Because x represents the number of calculators sold, we have x Ú 0, so
21,000 - 150p Ú 0. Solving this linear inequality gives p … 140. In addition,
Texas Instruments will charge only a positive price for the calculator, so p 7 0.
Combining these inequalities gives the domain of R, which is { p | 0 6 p … 140}.
(c) The function R is a quadratic function with a = - 150, b = 21,000, and c = 0.
Because a 6 0, the vertex is the highest point on the parabola. The revenue R is
a maximum when the price p is
p = -
21,000
b
= = +70.00
2a c
21 - 1502
a = - 150, b = 21,000
(d) The maximum revenue R is
R 1702 = - 1501702 2 + 21,0001702 = +735,000
(e) The number of calculators sold is given by the demand equation
x = 21,000 - 150p. At a price of p = +70,
x = 21,000 - 1501702 = 10,500
Revenue (dollars)
calculators are sold.
(f) To graph R, plot the intercept 1140, 02 and the vertex 170, 735 0002 . See Figure 23
for the graph.
R
800,000
700,000
600,000
500,000
400,000
300,000
200,000
100,000
(70, 735 000)
0
Figure 23
14 28 42 56 70 84 98 112 126 140
Price per calculator (dollars)
p
(g) Graph R = 675,000 and R 1p2 = - 150p2 + 21,000p on the same Cartesian
plane. See Figure 24. We find where the graphs intersect by solving
Revenue (dollars)
675,000
2
150p - 21,000p + 675,000
p2 - 140p + 4500
1p - 502 1p - 902
p = 50 or p
R
800,000
700,000
600,000
500,000
400,000
300,000
200,000
100,000
- 150p2 + 21,000p
0
Add 150p 2 - 21,000p to both sides.
Divide both sides by 150.
0
Factor.
0
90 Use the Zero-Product Property.
(70, 735 000)
(90, 675 000)
(50, 675 000)
0
Figure 24
=
=
=
=
=
14 28 42 56 70 84 98 112 126 140
Price per calculator (dollars)
p
304
CHAPTER 4 Linear and Quadratic Functions
The graphs intersect at 150, 675 0002 and 190, 675 0002 . Based on the graph
in Figure 24, Texas Instruments should charge between $50 and $90 to earn at
least $675,000 in revenue.
Now Work
EX AMPLE 2
PROBLEM
r
3
Maximizing the Area Enclosed by a Fence
A farmer has 2000 yards of fence to enclose a rectangular field. What are the dimensions
of the rectangle that encloses the most area?
Solution Figure 25 illustrates the situation. The available fence represents the
perimeter of the rectangle. If x is the length and w is the width, then
2x + 2w = 2000
(1)
The area A of the rectangle is
A = xw
To express A in terms of a single variable, solve equation (1) for w and substitute
the result in A = xw. Then A involves only the variable x. [You could also solve
equation (1) for x and express A in terms of w alone. Try it!]
x
w
w
x
Figure 25
2x + 2w = 2000
2w = 2000 - 2x
2000 - 2x
w =
= 1000 - x
2
Then the area A is
A = xw = x11000 - x2 = - x2 + 1000x
Now, A is a quadratic function of x.
A 1x2 = - x2 + 1000x
A
a = - 1, b = 1000, c = 0
Figure 26 shows the graph of A 1x2 = - x2 + 1000x. Because a 6 0, the vertex
is a maximum point on the graph of A. The maximum value occurs at
(500, 250 000)
x = -
250,000
b
1000
= = 500
2a
21 - 12
The maximum value of A is
(0, 0)
(1000, 0)
500
1000 x
Figure 26 A(x) = - x2 + 1000x
Aa -
b
b = A 15002 = - 5002 + 100015002 = - 250,000 + 500,000 = 250,000
2a
The largest rectangle that can be enclosed by 2000 yards of fence has an area of
250,000 square yards. Its dimensions are 500 yards by 500 yards.
Now Work
EX AMPLE 3
PROBLEM
r
7
Analyzing the Motion of a Projectile
A projectile is fired from a cliff 500 feet above the water at an inclination of 45°
to the horizontal, with a muzzle velocity of 400 feet per second. From physics, the
height h of the projectile above the water can be modeled by
h 1x2 =
- 32x2
+ x + 500
14002 2
where x is the horizontal distance of the projectile from the base of the cliff. See Figure 27.
h (x)
2500
2000
1500
1000
500
Figure 27
458
1000 2000
3000 4000
5000
x
SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data
305
(a) Find the maximum height of the projectile.
(b) How far from the base of the cliff will the projectile strike the water?
Solution
(a) The height of the projectile is given by a quadratic function.
h 1x2 =
- 32x2
-1 2
+ x + 500 =
x + x + 500
2
5000
14002
We are looking for the maximum value of h. Because a 6 0, the maximum value
occurs at the vertex, whose x-coordinate is
x = -
b
= 2a
1
5000
=
= 2500
-1
2
2a
b
5000
The maximum height of the projectile is
h 125002 =
-1
125002 2 + 2500 + 500 = - 1250 + 2500 + 500 = 1750 ft
5000
(b) The projectile will strike the water when the height is zero. To find the distance x
traveled, solve the equation
h 1x2 =
Seeing the Concept
-1 2
x + x + 500 = 0
5000
The discriminant of this quadratic equation is
Graph
-1 2
h(x) =
x + x + 500
5000
0 … x … 5500
Use MAXIMUM to find the maximum
height of the projectile, and use ROOT
or ZERO to find the distance from the
base of the cliff to where it strikes the
water. Compare your results with those
obtained in Example 3.
b2 - 4ac = 12 - 4a
Then
x =
- b { 2b2 - 4ac
- 1 { 21.4
- 458
=
≈ e
2a
5458
-1
2a
b
5000
Discard the negative solution. The projectile will strike the water at a distance
of about 5458 feet from the base of the cliff.
r
Now Work
EXAMPL E 4
-1
b 15002 = 1.4
5000
PROBLEM
11
The Golden Gate Bridge
The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco
Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two
huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above
the water. The cables are parabolic in shape* and touch the road surface at the
center of the bridge. Find the height of the cable above the road at a distance of
1000 feet from the center.
Solution
See Figure 28 on page 306. Begin by choosing the placement of the coordinate
axes so that the x-axis coincides with the road surface and the origin coincides with
the center of the bridge. As a result, the 746-foot towers will be vertical (height
746 - 220 = 526 feet above the road) and located 2100 feet from the center. Also,
the cable, which has the shape of a parabola, will extend from the towers, open up,
and have its vertex at 10, 02. This choice of placement of the axes enables the equation
of the parabola to have the form y = ax2, a 7 0. Note that the points 1 - 2100, 5262
and 12100, 5262 are on the graph.
*A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is
suspended from the cable, the cable takes the shape of a parabola.
CHAPTER 4 Linear and Quadratic Functions
(2100, 526)
(2100, 526)
y
526'
306
(0, 0)
2100'
746'
x
1000'
220'
Figure 28
?
2100'
Use these facts to find the value of a in y = ax2.
y = ax2
526 = a121002 2
a =
x = 2100, y = 526
526
121002 2
The equation of the parabola is
y =
526
x2
121002 2
When x = 1000, the height of the cable is
y =
526
110002 2 ≈ 119.3 feet
121002 2
The cable is 119.3 feet above the road at a distance of 1000 feet from the center of
the bridge.
r
Now Work
PROBLEM
13
2 Build Quadratic Models from Data
In Section 4.2, we found the line of best fit for data that appeared to be linearly
related. It was noted that data may also follow a nonlinear relation. Figures 29(a)
and (b) show scatter diagrams of data that follow a quadratic relation.
y ax 2 bx c, a 0
Figure 29
E X AMPLE 5
(a)
y ax 2 bx c, a 0
(b)
Fitting a Quadratic Function to Data
The data in Table 8 on page 307 represent the percentage D of the population that is
divorced for various ages x in 2012.
(a) Draw a scatter diagram of the data, treating age as the independent variable.
Comment on the type of relation that may exist between age and percentage of
the population divorced.
(b) Use a graphing utility to find the quadratic function of best fit that models the
relation between age and percentage of the population divorced.
(c) Use the model found in part (b) to approximate the age at which the percentage
of the population divorced is greatest.
(d) Use the model found in part (b) to approximate the highest percentage of the
population that is divorced.
(e) Use a graphing utility to draw the quadratic function of best fit on the scatter diagram.
307
SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data
Table 8
Age, x
Percentage Divorced, D
22
0.9
27
3.6
32
7.4
37
10.4
42
12.7
50
15.7
60
16.2
70
13.1
80
6.5
Source: United States Statistical Abstract, 2012
Solution
19
15
0
(a) Figure 30 shows the scatter diagram, from which it appears the data follow a
quadratic relation, with a 6 0.
(b) Execute the QUADratic REGression program to obtain the results shown in
Figure 31. The output shows the equation y = ax2 + bx + c. The quadratic
function of best fit that models the relation between age and percentage
divorced is
D1x2 = - 0.0143x2 + 1.5861x - 28.1886
85
The model
where x represents age and D represents the percentage divorced.
(c) Based on the quadratic function of best fit, the age with the greatest percentage
divorced is
Figure 30
-
b
1.5861
= ≈ 55 years
2a
21 - 0.01432
(d) Evaluate the function D1x2 at x = 55.
D1552 = - 0.01431552 2 + 1.58611552 - 28.1886 ≈ 15.8 percent
According to the model, 55-year-olds have the highest percentage divorced at
15.8 percent.
(e) Figure 32 shows the graph of the quadratic function found in part (b) drawn on
the scatter diagram.
Figure 31
r
19
15
0
85
Look again at Figure 31. Notice that the output given by the graphing calculator
does not include r, the correlation coefficient. Recall that the correlation coefficient
is a measure of the strength of a linear relation that exists between two variables.
The graphing calculator does not provide an indication of how well the function
fits the data in terms of r, since a quadratic function cannot be expressed as a linear
function.
Figure 32
Now Work
PROBLEM
25
4.4 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Translate the following sentence into a mathematical equation:
The total revenue R from selling x hot dogs is $3 times the
number of hot dogs sold. (pp. 134–140)
2. Use a graphing utility to find the line of best fit for the
following data: (pp. 284–287)
x
3
5
5
6
7
8
y
10
13
12
15
16
19
308
CHAPTER 4 Linear and Quadratic Functions
Applications and Extensions
3. Maximizing Revenue The price p (in dollars) and the
quantity x sold of a certain product obey the demand equation
1
x + 100
6
Find a model that expresses the revenue R as a function
of x. (Remember, R = xp.)
What is the domain of R?
What is the revenue if 200 units are sold?
What quantity x maximizes revenue? What is the
maximum revenue?
What price should the company charge to maximize
revenue?
p = -
(a)
(b)
(c)
(d)
(e)
4. Maximizing Revenue The price p (in dollars) and the
quantity x sold of a certain product obey the demand equation
1
x + 100
3
Find a model that expresses the revenue R as a function
of x.
What is the domain of R?
What is the revenue if 100 units are sold?
What quantity x maximizes revenue? What is the
maximum revenue?
What price should the company charge to maximize
revenue?
9. Enclosing the Most Area with a Fence A farmer with 4000
meters of fencing wants to enclose a rectangular plot that
borders on a river. If the farmer does not fence the side
along the river, what is the largest area that can be enclosed?
(See the figure.)
x
x
p = -
(a)
(b)
(c)
(d)
(e)
5. Maximizing Revenue The price p (in dollars) and the
quantity x sold of a certain product obey the demand equation
x = - 5p + 100
0 6 p … 20
(a) Express the revenue R as a function of x.
(b) What is the revenue if 15 units are sold?
(c) What quantity x maximizes revenue? What is the
maximum revenue?
(d) What price should the company charge to maximize
revenue?
(e) What price should the company charge to earn at least
$480 in revenue?
6. Maximizing Revenue The price p (in dollars) and the
quantity x sold of a certain product obey the demand equation
x = - 20p + 500
0 6 p … 25
(a) Express the revenue R as a function of x.
(b) What is the revenue if 20 units are sold?
(c) What quantity x maximizes revenue? What is the
maximum revenue?
(d) What price should the company charge to maximize
revenue?
(e) What price should the company charge to earn at least
$3000 in revenue?
7. Enclosing a Rectangular Field David has 400 yards of fencing
and wishes to enclose a rectangular area.
(a) Express the area A of the rectangle as a function of the
width w of the rectangle.
(b) For what value of w is the area largest?
(c) What is the maximum area?
8. Enclosing a Rectangular Field Beth has 3000 feet of fencing
available to enclose a rectangular field.
(a) Express the area A of the rectangle as a function of x,
where x is the length of the rectangle.
(b) For what value of x is the area largest?
(c) What is the maximum area?
4000 2x
10. Enclosing the Most Area with a Fence A farmer with 2000
meters of fencing wants to enclose a rectangular plot that
borders on a straight highway. If the farmer does not fence
the side along the highway, what is the largest area that can
be enclosed?
11. Analyzing the Motion of a Projectile A projectile is fired
from a cliff 200 feet above the water at an inclination of
45° to the horizontal, with a muzzle velocity of 50 feet per
second. The height h of the projectile above the water is
modeled by
h1x2 =
- 32x2
+ x + 200
1502 2
where x is the horizontal distance of the projectile from the
face of the cliff.
(a) At what horizontal distance from the face of the cliff is
the height of the projectile a maximum?
(b) Find the maximum height of the projectile.
(c) At what horizontal distance from the face of the cliff
will the projectile strike the water?
(d) Using a graphing utility, graph the function h,
0 … x … 200.
(e) Use a graphing utility to verify the solutions found in
parts (b) and (c).
(f) When the height of the projectile is 100 feet above the
water, how far is it from the cliff?
12. Analyzing the Motion of a Projectile A projectile is fired
at an inclination of 45° to the horizontal, with a muzzle
velocity of 100 feet per second. The height h of the projectile is
modeled by
h1x2 =
- 32x2
+ x
11002 2
where x is the horizontal distance of the projectile from the
firing point.
(a) At what horizontal distance from the firing point is the
height of the projectile a maximum?
(b) Find the maximum height of the projectile.
(c) At what horizontal distance from the firing point will
the projectile strike the ground?
(d) Using a graphing utility, graph the function h,
0 … x … 350.
SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data
309
(e) Use a graphing utility to verify the results obtained in
parts (b) and (c).
(f) When the height of the projectile is 50 feet above the ground,
how far has it traveled horizontally?
13. Suspension Bridge A suspension bridge with weight
uniformly distributed along its length has twin towers that
extend 75 meters above the road surface and are 400 meters
apart. The cables are parabolic in shape and are suspended
from the tops of the towers. The cables touch the road surface
at the center of the bridge. Find the height of the cables at a
point 100 meters from the center. (Assume that the road is
level.)
14. Architecture A parabolic arch has a span of 120 feet and
a maximum height of 25 feet. Choose suitable rectangular
coordinate axes and find the equation of the parabola. Then
calculate the height of the arch at points 10 feet, 20 feet, and
40 feet from the center.
15. Constructing Rain Gutters A rain gutter is to be made of
aluminum sheets that are 12 inches wide by turning up the
edges 90°. See the illustration.
(a) What depth will provide maximum cross-sectional area
and hence allow the most water to flow?
(b) What depths will allow at least 16 square inches of water
to flow?
18. Architecture A special window has the shape of a rectangle
surmounted by an equilateral triangle. See the figure. If the
perimeter of the window is 16 feet, what dimensions will
admit the most light?
13 2
[Hint: Area of an equilateral triangle = a
bx , where x
4
is the length of a side of the triangle.]
x
x
x
19. Chemical Reactions A self-catalytic chemical reaction
results in the formation of a compound that causes the
formation ratio to increase. If the reaction rate V is modeled by
x
x
12
12 2x
in.
x
x
V 1x2 = kx1a - x2,
0 … x … a
where k is a positive constant, a is the initial amount of the
compound, and x is the variable amount of the compound,
for what value of x is the reaction rate a maximum?
20. Calculus: Simpson’s Rule The figure shows the graph
of y = ax2 + bx + c. Suppose that the points 1 - h, y0 2,
10, y1 2, and 1h, y2 2 are on the graph. It can be shown that
the area enclosed by the parabola, the x-axis, and the lines
x = - h and x = h is
Area =
h
12ah2 + 6c2
3
Show that this area may also be given by
16. Norman Windows A Norman window has the shape of a
rectangle surmounted by a semicircle of diameter equal to
the width of the rectangle. See the figure. If the perimeter of
the window is 20 feet, what dimensions will admit the most
light (maximize the area)?
[Hint: Circumference of a circle = 2pr; area of a
circle = pr 2, where r is the radius of the circle.]
Area =
h
1y + 4y1 + y2 2
3 0
y
(0, y1)
(h, y2)
(h, y0)
h
h
x
21. Use the result obtained in Problem 20 to find the area
enclosed by f 1x2 = - 5x2 + 8, the x-axis, and the lines
x = - 1 and x = 1.
22. Use the result obtained in Problem 20 to find the area
enclosed by f 1x2 = 2x2 + 8, the x-axis, and the lines
x = - 2 and x = 2.
17. Constructing a Stadium A track-and-field playing area is in
the shape of a rectangle with semicircles at each end. See the
figure (top, right). The inside perimeter of the track is to be
1500 meters. What should the dimensions of the rectangle be
so that the area of the rectangle is a maximum?
23. Use the result obtained in Problem 20 to find the area
enclosed by f 1x2 = x2 + 3x + 5, the x-axis, and the lines
x = - 4 and x = 4.
24. Use the result obtained in Problem 20 to find the area
enclosed by f 1x2 = - x2 + x + 4, the x-axis, and the lines
x = - 1 and x = 1.
310
CHAPTER 4 Linear and Quadratic Functions
25. Life Cycle Hypothesis An individual’s income varies with
his or her age. The following table shows the median income I
of males of different age groups within the United States
for 2012. For each age group, let the class midpoint represent
the independent variable, x. For the class “65 years and
older,” we will assume that the class midpoint is 69.5.
Class
Midpoint, x
Median
Income, I
15–24 years
19.5
$10,869
25–34 years
29.5
$34,113
35–44 years
39.5
$45,225
45–54 years
49.5
$46,466
55–64 years
59.5
$42,176
65 years and older
69.5
$27,612
Age
Source: U.S. Census Bureau
(a) Use a graphing utility to draw a scatter diagram of the
data. Comment on the type of relation that may exist
between the two variables.
(b) Use a graphing utility to find the quadratic function of
best fit that models the relation between age and median
income.
(c) Use the function found in part (b) to determine the age at
which an individual can expect to earn the most income.
(d) Use the function found in part (b) to predict the peak
income earned.
(e) With a graphing utility, graph the quadratic function of
best fit on the scatter diagram.
26. Height of a Ball A shot-putter throws a ball at an inclination
of 45° to the horizontal. The following data represent the height
of the ball h, in feet, at the instant that it has traveled x feet
horizontally.
Distance, x
Height, h
20
25
40
40
60
55
80
65
100
71
120
77
140
77
160
75
180
71
200
64
(a) Use a graphing utility to draw a scatter diagram of the
data. Comment on the type of relation that may exist
between the two variables.
(b) Use a graphing utility to find the quadratic function of
best fit that models the relation between distance and
height.
(c) Use the function found in part (b) to determine how far
the ball will travel before it reaches its maximum height.
(d) Use the function found in part (b) to find the maximum
height of the ball.
(e) With a graphing utility, graph the quadratic function of
best fit on the scatter diagram.
Mixed Practice
27. Which Model? The following data represent the square
footage and rents (dollars per month) for apartments in
the La Jolla area of San Diego, California.
Square Footage, x
Rent per Month, R
520
$1525
621
$1750
718
$1785
753
$1850
850
$1900
968
$2130
1020
$2180
Source: apartments.com, 2014
(a) Using a graphing utility, draw a scatter diagram of
the data treating square footage as the independent
variable. What type of relation appears to exist
between square footage and rent?
(b) Based on your response to part (a), find either a linear or
a quadratic model that describes the relation between
square footage and rent.
(c) Use your model to predict the rent for an apartment in
San Diego that is 875 square feet.
28. Which Model? An engineer collects the following data
showing the speed s of a Toyota Camry and its average
miles per gallon, M.
Speed, s
Miles per Gallon, M
30
18
35
20
40
23
40
25
45
25
50
28
55
30
60
29
65
26
65
25
70
25
(a) Using a graphing utility, draw a scatter diagram of the
data, treating speed as the independent variable. What
type of relation appears to exist between speed and
miles per gallon?
SECTION 4.4 Build Quadratic Models from Verbal Descriptions and from Data
(b) Based on your response to part (a), find either a linear
model or a quadratic model that describes the relation
between speed and miles per gallon.
(c) Use your model to predict the miles per gallon for a
Camry that is traveling 63 miles per hour.
29. Which Model? The following data represent the birth
rate (births per 1000 population) for women whose age is
a, in 2012.
Age, a
30. Which Model? A cricket makes a chirping noise by sliding
its wings together rapidly. Perhaps you have noticed that the
number of chirps seems to increase with the temperature.
The following data list the temperature (in degrees
Fahrenheit) and the number of chirps per second for the
striped ground cricket.
Birth Rate, B
16
14.1
19
51.4
22
83.1
27
106.5
32
97.3
37
48.3
42
10.4
Temperature (ºF), x
Chirps per Second, C
88.6
20.0
93.3
19.8
80.6
17.1
69.7
14.7
69.4
15.4
79.6
15.0
80.6
16.0
76.3
14.4
75.2
15.5
Source: Pierce, George W. The Songs of Insects. Cambridge,
MA Harvard University Press, 1949, pp. 12 – 21
Source: National Vital Statistics
System, 2013
(a) Using a graphing utility, draw a scatter diagram of the
data, treating age as the independent variable. What type
of relation appears to exist between age and birth rate?
(b) Based on your response to part (a), find either a linear or
a quadratic model that describes the relation between age
and birth rate.
(c) Use your model to predict the birth rate for 35-yearold women.
311
(a) Using a graphing utility, draw a scatter diagram of
the data, treating temperature as the independent
variable. What type of relation appears to exist between
temperature and chirps per second?
(b) Based on your response to part (a), find either a linear
or a quadratic model that best describes the relation
between temperature and chirps per second.
(c) Use your model to predict the chirps per second if the
temperature is 80°F.
Explaining Concepts: Discussion and Writing
31. Refer to Example 1 in this section. Notice that if the price
charged for the calculators is $0 or $140, then the revenue
is $0. It is easy to explain why revenue would be $0 if the
price charged were $0, but how can revenue be $0 if the price
charged is $140?
Retain Your Knowledge
Problems 32–35 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
32. Express as a complex number: 2- 225
33. Find the distance between the points P1 = (4, - 7) and P2 = ( - 1, 5).
34. Find the equation of the circle with center (– 6, 0) and radius r = 27.
35. Solve: 5x2 + 8x - 3 = 0
‘Are You Prepared?’ Answers
1. R = 3x
2. y = 1.7826x + 4.0652
312
CHAPTER 4 Linear and Quadratic Functions
4.5 Inequalities Involving Quadratic Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
r Solve Inequalities (Section 1.5, pp. 123–126)
r Use Interval Notation (Section 1.5, pp. 120–121)
Now Work the ‘Are You Prepared?’ problems on page 314.
OBJECTIVE 1 Solve Inequalities Involving a Quadratic Function (p. 312)
1 Solve Inequalities Involving a Quadratic Function
In this section we solve inequalities that involve quadratic functions. We will accomplish
this by using their graphs. For example, to solve the inequality
ax2 + bx + c 7 0
a ≠ 0
graph the function f1x2 = ax2 + bx + c and, from the graph, determine
where it is above the x-axis—that is, where f1x2 7 0. To solve the inequality
ax2 + bx + c 6 0, a ≠ 0, graph the function f 1x2 = ax2 + bx + c and determine
where the graph is below the x-axis. If the inequality is not strict, include the
x-intercepts, if any, in the solution.
Solving an Inequality
EX AMPLE 1
Solve the inequality x2 - 4x - 12 … 0 and graph the solution set.
Solution
y
8
Graph the function f 1x2 = x2 - 4x - 12.
f102 = - 12
y-intercept:
x-intercepts (if any):
4
(6, 0)
(–2, 0)
–4
8 x
4
x - 4x - 12 = 0
2
1x - 62 1x + 22 = 0
x - 6 = 0 or x + 2 = 0
x = 6 or
x = -2
–4
Evaluate f at 0.
Solve f1x2 = 0.
Factor.
Apply the Zero-Product Property.
–8
The y-intercept is - 12; the x-intercepts are - 2 and 6.
b
-4
= = 2. Because f122 = - 16, the vertex is
The vertex is at x = 2a
2
at 12, - 162 .
–12 (0, –12)
–16
(2, –16)
Figure 33 f (x) = x2 - 4x - 12
–4
–2
0
2
4
6
8
x
See Figure 33 for the graph.
The graph is below the x-axis for - 2 6 x 6 6. Because the original inequality
is not strict, include the x-intercepts. The solution set is 5 x 0 - 2 … x … 66 or, using
interval notation, 3 - 2, 64 . See Figure 34 for the graph of the solution set.
r
Figure 34
Now Work
EX AMPLE 2
PROBLEM
9
Solving an Inequality
Solve the inequality 2x2 6 x + 10 and graph the solution set.
Solution
Option 1
Rearrange the inequality so that 0 is on the right side.
2x2 6 x + 10
Subtract x + 10 from both sides.
2x2 - x - 10 6 0
This inequality is equivalent to the original inequality.
SECTION 4.5 Inequalities Involving Quadratic Functions
Next graph the function f1x2 = 2x2 - x - 10 to find where f1x2 6 0.
y
4
f102 = - 10
y-intercept:
2
(2, 0)
2
2
x-intercepts (if any):
4
x
6
( 1–4 , 10.125)
Figure 35 f (x) = 2x2 - x - 10
g(x) x 10
y
12
–– )
( 5–2 , 25
2
10
Option 2 If f1x2 = 2x2 and g1x2 = x + 10, then the inequality to be solved is
f1x2 6 g1x2. Graph the functions f1x2 = 2x2 and g1x2 = x + 10. See Figure 36.
The graphs intersect where f1x2 = g1x2. Then
2x2 - x - 10 = 0
f(x) 2x 2
12x - 52 1x + 22 = 0
4
2
4
x =
x
Figure 36
2
0
2
4
EXAM PLE 3
y
3
PROBLEMS
5
AND
13
Solving an Inequality
Solve the inequality x2 + x + 1 7 0 and graph the solution set.
(1, 3)
Solution
2
(0, 1)
–2 –1
1
2
x
( 1–2, 3–4)
Figure 38 f 1x2 = x2 + x + 1
2
Figure 39
x = -2
or
r
Now Work
4
5
2
Apply the Zero-Product Property.
5 25
The graphs intersect at the points 1 - 2, 82 and a , b . To solve f 1x2 6 g1x2, find
2 2
where the graph of f is below the graph of g. This happens between the points of
5
intersection. Because the inequality is strict, the solution set is e x ` - 2 6 x 6 f or,
2
5
using interval notation, a - 2, b .
2
See Figure 37 for the graph of the solution set.
Figure 37
(–1, 1)
Factor.
2x - 5 = 0 or x + 2 = 0
2
4
f1x2 = g1x2
2x2 = x + 10
2
Apply the Zero-Product
Property.
5
The y-intercept is - 10; the x-intercepts are - 2 and .
2
b
-1
1
1
The vertex is at x = = = . Because f a b = - 10.125, the vertex
2a
4
4
4
1
is a , - 10.125b . See Figure 35 for the graph.
4
5
The graph is below the x-axis 1f1x2 6 02 between x = - 2 and x = . Because
2
5
the inequality is strict, the solution set is e x ` - 2 6 x 6 f or, using interval
2
5
notation, a - 2, b .
2
8
6
4
Solve f1x2 = 0.
Factor.
2x - 5 = 0 or x + 2 = 0
5
x =
or x = - 2
2
2
(2, 8)
Evaluate f at 0.
2x - x - 10 = 0
12x - 52 1x + 22 = 0
( 5–2 , 0)
4
10
313
0
2
Graph the
x-intercepts
b
x = =
2a
and 1 - 1, 12
function f 1x2 = x2 + x + 1. The y-intercept is 1; there are no
(Do you see why? Check the discriminant). The vertex is at
1
1
3
1 3
- . Since f a - b = , the vertex is at a - , b . The points 11, 32
2
2
4
2 4
are also on the graph. See Figure 38.
The graph of f lies above the x-axis for all x. The solution set is the set of all real
numbers, or ( - q , q ). See Figure 39.
r
4
Now Work
PROBLEM
17
314
CHAPTER 4 Linear and Quadratic Functions
4.5 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
2. Write 1 - 2, 7] using inequality notation. (pp. 120–121)
1. Solve the inequality - 3x - 2 6 7 (pp. 123–126)
Skill Building
In Problems 3–6, use the figure to solve each inequality.
3.
y
y f(x)
4.
y
5
5.
(1.5, 5)
(2, 8)
3
(2, 0)
y
6.
y f(x)
8
3
3
x
4
3
(3, 12)
(4, 0)
2
2
4
x
4
(1, 2)
4
5
3
x
(1, 3)
12
4
(3, 12)
y f(x)
x
y g(x)
4
(a) f 1x2 7 0
(b) f 1x2 … 0
y g(x)
6
(2, 8)
(2, 0)
(1, 0)
y
y g(x)
(a) g1x2 Ú f 1x2
(b) f 1x2 7 g1x2
(a) g1x2 6 0
(b) g1x2 Ú 0
(a) f 1x2 6 g1x2
(b) f 1x2 Ú g1x2
In Problems 7–22, solve each inequality.
7.
11.
15.
19.
x2 - 3x - 10 6 0
x2 - 9 6 0
2x2 6 5x + 3
4x2 + 9 6 6x
8.
12.
16.
20.
x2 + 3x - 10 7 0
x2 - 1 6 0
6x2 6 6 + 5x
25x2 + 16 6 40x
9.
13.
17.
21.
x2 - 4x 7 0
x2 + x 7 12
x2 - x + 1 … 0
61x2 - 12 7 5x
10.
14.
18.
22.
x2 + 8x
x2 + 7x
x2 + 2x
212x2 -
7 0
6 - 12
+ 4 7 0
3x2 7 - 9
Mixed Practice
23. What is the domain of the function f 1x2 = 2x2 - 16?
In Problems 25–32, use the given functions f and g.
(a) Solve f 1x2 = 0.
(e) Solve g1x2 … 0.
25. f 1x2 = x2 - 1
g1x2 = 3x + 3
29. f 1x2 = x2 - 4
g1x2 = - x2 + 4
(b) Solve g1x2 = 0.
24. What is the domain of the function f 1x2 = 2x - 3x2?
(c) Solve f 1x2 = g1x2.
(d) Solve f 1x2 7 0.
27. f 1x2 = - x2 + 1
g1x2 = 4x + 1
28. f 1x2 = - x2 + 4
g1x2 = - x - 2
(f) Solve f 1x2 7 g1x2.
(g) Solve f 1x2 Ú 1.
30. f 1x2 = x2 - 2x + 1
g1x2 = - x2 + 1
31. f 1x2 = x2 - x - 2
g1x2 = x2 + x - 2
26. f 1x2 = - x2 + 3
g1x2 = - 3x + 3
32. f 1x2 = - x2 - x + 1
g1x2 = - x2 + x + 6
Applications and Extensions
33. Physics A ball is thrown vertically upward with an initial
velocity of 80 feet per second. The distance s (in feet) of the
ball from the ground after t seconds is s 1t2 = 80t - 16t 2.
96 ft
s 80t 16t 2
(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 96 feet above the
ground?
34. Physics A ball is thrown vertically upward with an initial
velocity of 96 feet per second. The distance s (in feet) of the
ball from the ground after t seconds is s 1t2 = 96t - 16t 2.
(a) At what time t will the ball strike the ground?
(b) For what time t is the ball more than 128 feet above the
ground?
35. Revenue Suppose that the manufacturer of a gas clothes
dryer has found that when the unit price is p dollars, the
revenue R (in dollars) is
R 1p2 = - 4p2 + 4000p
Chapter Review
(a) At what prices p is revenue zero?
(b) For what range of prices will revenue exceed $800,000?
(a) If the round must clear a hill 200 meters high at a
distance of 2000 meters in front of the howitzer, what c
values are permitted in the trajectory equation?
(b) If the goal in part (a) is to hit a target on the ground
75 kilometers away, is it possible to do so? If so, for what
values of c? If not, what is the maximum distance the
round will travel?
Source: www.answers.com
36. Revenue The John Deere company has found that the
revenue from sales of heavy-duty tractors is a function of
the unit price p, in dollars, that it charges. The revenue R, in
dollars, is given by
R 1p2 = -
1 2
p + 1900p
2
(a) At what prices p is revenue zero?
(b) For what range of prices will revenue exceed
$1,200,000?
37. Artillery A projectile fired from the point (0, 0) at an angle
to the positive x-axis has a trajectory given by
g x 2
y = cx - 11 + c 2 2 a b a b
2 v
where
x =
y =
v =
g =
horizontal distance in meters
height in meters
initial muzzle velocity in meters per second (m/sec)
acceleration due to gravity = 9.81 meters per
second squared (m/sec2)
c 7 0 is a constant determined by the angle of elevation.
A howitzer fires an artillery round with a muzzle velocity
of 897 m/sec.
315
38. Runaway Car Using Hooke’s Law, we can show that the
work done in compressing a spring a distance of x feet from its
1
at-rest position is W = kx2, where k is a stiffness constant
2
depending on the spring. It can also be shown that the work
done by a body in motion before it comes to rest is given
w 2
∼
v , where w = weight of the object (in lbs),
by W =
2g
g = acceleration due to gravity (32.2 ft/sec2), and v = object’s
velocity (in ft/sec). A parking garage has a spring shock
absorber at the end of a ramp to stop runaway cars. The
spring has a stiffness constant k = 9450 lb/ft and must be able
to stop a 4000-lb car traveling at 25 mph. What is the least
compression required of the spring? Express your answer
using feet to the nearest tenth.
∼
[Hint: Solve W 7 W, x Ú 0].
Source: www.sciforums.com
Explaining Concepts: Discussion and Writing
39. Show that the inequality 1x - 42 2 … 0 has exactly one solution.
40. Show that the inequality 1x - 22 7 0 has one real number
that is not a solution.
2
2
41. Explain why the inequality x + x + 1 7 0 has all real
numbers as the solution set.
42. Explain why the inequality x2 - x + 1 6 0 has the empty
set as the solution set.
43. Explain the circumstances under which the x-intercepts
of the graph of a quadratic function are included in the
solution set of a quadratic inequality.
Retain Your Knowledge
Problems 44–47 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
2
44. Determine the domain of f(x) = 210 - 2x.
46. Consider the linear function f(x) = x - 6.
3
-x
is even,
45. Determine algebraically whether f(x) = 2
(a) Find the intercepts of the graph of f.
x + 9
odd, or neither.
(b) Graph f.
47. Multiply (6 - 5i)(4 - i). Write the answer in the form
a + bi.
‘Are You Prepared?’ Answers
1. 5x 0 x 7 - 36 or 1 - 3, q 2
2. - 2 6 x … 7
Chapter Review
Things to Know
Linear function (p. 274)
f 1x2 = mx + b
Average rate of change = m
The graph is a line with slope m and y-intercept b.
316
CHAPTER 4 Linear and Quadratic Functions
Quadratic function (pp. 291–295)
f 1x2 = ax2 + bx + c, a ≠ 0
The graph is a parabola that opens up if a 7 0 and opens down if a 6 0.
Vertex: a -
b
b
, f a- b b
2a
2a
Axis of symmetry: x = y-intercept: f 102 = c
b
2a
x-intercept(s): If any, found by finding the real solutions of the equation ax2 + bx + c = 0
Objectives
Section
4.1
You should be able to…
Examples
Review Exercises
Graph linear functions (p. 274)
Use average rate of change to identify linear functions (p. 274)
Determine whether a linear function is increasing, decreasing, or constant (p. 277)
Build linear models from verbal descriptions (p. 278)
1
2
3
4, 5
1(a)–3(a), 1(c)–3(c)
1(b)–3(b), 4, 5
1(d)–3(d)
21
Draw and interpret scatter diagrams (p. 284)
Distinguish between linear and nonlinear relations (p. 285)
Use a graphing utility to find the line of best fit (p. 286)
1
2
4
29(a), 30(a)
29(b), 30(a)
29(c)
Graph a quadratic function using transformations (p. 292)
Identify the vertex and axis of symmetry of a quadratic function (p. 294)
Graph a quadratic function using its vertex, axis, and intercepts (p. 294)
Find a quadratic function given its vertex and one other point (p. 297)
Find the maximum or minimum value of a quadratic function (p. 298)
1
2
3–5
6
7
6–8
9–13
9–13
19, 20
14–16, 22–27
2
Build quadratic models from verbal descriptions (p. 302)
Build quadratic models from data (p. 306)
1–4
5
22–28
30
3
Solve inequalities involving a quadratic function (p. 312)
1–3
17, 18
1
2
3
4
4.2
1
2
3
4.3
1
2
3
4
5
4.4
1
4.5
Review Exercises
In Problems 1–3:
(a) Determine the slope and y-intercept of each linear function.
(b) Find the average rate of change of each function.
(c) Graph each function. Label the intercepts.
(d) Determine whether the function is increasing, decreasing, or constant.
4
1. f 1x2 = 2x - 5
2. h1x2 = x - 6
5
3. G1x2 = 4
In Problems 4 and 5, determine whether the function is linear or nonlinear. If the function is linear, state its slope.
4.
x
y = f(x)
-1
5.
x
y = g(x)
-2
-1
-3
0
3
0
4
1
8
1
7
2
13
2
6
3
18
3
1
In Problems 6–8, graph each quadratic function using transformations (shifting, compressing, stretching, and/or reflecting).
6. f 1x2 = 1x + 12 2 - 4
7. f 1x2 = - 1x - 42 2
8. f 1x2 = - 31x + 22 2 + 1
Chapter Review
317
In Problems 9–13, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of
symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function
is increasing and where it is decreasing.
1
9. f 1x2 = 1x - 22 2 + 2
10. f 1x2 = x2 - 16
11. f 1x2 = - 4x2 + 4x
4
9
12. f 1x2 = x2 + 3x + 1
13. f 1x2 = 3x2 + 4x - 1
2
In Problems 14–16, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value.
15. f 1x2 = - x2 + 8x - 4
14. f 1x2 = 3x2 - 6x + 4
16. f 1x2 = - 3x2 + 12x + 4
In Problems 17 and 18, solve each quadratic inequality.
17. x2 + 6x - 16 6 0
18. 3x2 Ú 14x + 5
In Problems 19 and 20, find the quadratic function for which:
19. Vertex is (2, −4); y-intercept is −16
20. Vertex is 1 - 1, 22; contains the point 11, 62
21. Sales Commissions Bill was just offered a sales position for
a computer company. His salary would be $25,000 per year
plus 1% of his total annual sales.
(a) Find a linear function that relates Bill’s annual salary, S,
to his total annual sales, x.
(b) If Bill’s total annual sales were $1,000,000, what would
be Bill’s salary?
(c) What would Bill have to sell to earn $100,000?
(d) Determine the sales required of Bill for his salary to
exceed $150,000.
22. Demand Equation The price p (in dollars) and the quantity x
sold of a certain product obey the demand equation
p = -
1
x + 150
10
0 … x … 1500
(a) Express the revenue R as a function of x.
(b) What is the revenue if 100 units are sold?
(c) What quantity x maximizes revenue? What is the
maximum revenue?
(d) What price should the company charge to maximize
revenue?
25. Architecture A special window in the shape of a rectangle
with semicircles at each end is to be constructed so that the
outside perimeter is 100 feet. See the illustration. Find the
dimensions of the rectangle that maximizes its area.
26. Minimizing Marginal Cost Callaway Golf Company has
determined that the marginal cost C of manufacturing x
Big Bertha golf clubs may be expressed by the quadratic
function
C 1x2 = 4.9x2 - 617.4x + 19,600
(a) How many clubs should be manufactured to minimize
the marginal cost?
(b) At this level of production, what is the marginal cost?
27. Maximizing Area A rectangle has one vertex on the line
y = 10 - x, x 7 0, another at the origin, one on the positive
x-axis, and one on the positive y-axis. Express the area A of the
rectangle as a function of x. Find the largest area A that can
be enclosed by the rectangle.
28. Parabolic Arch Bridge A horizontal bridge is in the shape
of a parabolic arch. Given the information shown in the figure,
what is the height h of the arch 2 feet from shore?
23. Landscaping A landscape engineer has 200 feet of border
to enclose a rectangular pond. What dimensions will result
in the largest pond?
24. Enclosing the Most Area with a Fence A farmer with
10,000 meters of fencing wants to enclose a rectangular field
and then divide it into two plots with a fence parallel to one
of the sides. See the figure. What is the largest area that can
be enclosed?
10 ft
h
2 ft
20 ft
29. Bone Length Research performed at NASA, led by
Dr. Emily R. Morey-Holton, measured the lengths of the
right humerus and right tibia in 11 rats that were sent to
space on Spacelab Life Sciences 2. The data on page 318 were
collected.
(a) Draw a scatter diagram of the data, treating length of
the right humerus as the independent variable.
(b) Based on the scatter diagram, do you think that there
is a linear relation between the length of the right
humerus and the length of the right tibia?
318
CHAPTER 4 Linear and Quadratic Functions
(c) Use a graphing utility to find the line of best fit
relating length of the right humerus and length of the
right tibia.
(d) Predict the length of the right tibia on a rat whose right
humerus is 26.5 millimeters (mm).
(a) Draw a scatter diagram of the data. Comment on the
type of relation that may exist between the two variables.
Advertising
Expenditures ($1000s)
Total Revenue
($1000s)
20
6101
6222
Right Humerus
(mm), x
Right Tibia
(mm), y
22
25
6350
24.80
36.05
25
6378
24.59
35.57
27
6453
24.59
35.57
28
6423
24.29
34.58
29
6360
23.81
34.20
31
6231
24.87
34.73
25.90
37.38
26.11
37.96
26.63
37.46
26.31
37.75
26.84
38.50
(b) The quadratic function of best fit to these data is
R 1A2 = - 7.76A2 + 411.88A + 942.72
Source: NASA Life Sciences Data Archive
30. Advertising A small manufacturing firm collected the
following data on advertising expenditures A (in thousands
of dollars) and total revenue R (in thousands of dollars).
Chapter Test
Use this function to determine the optimal level of
advertising.
(c) Use the function to predict the total revenue when the
optimal level of advertising is spent.
(d) Use a graphing utility to verify that the function given in
part (b) is the quadratic function of best fit.
(e) Use a graphing utility to draw a scatter diagram of the
data, and then graph the quadratic function of best fit on
the scatter diagram.
The Chapter Test Prep Videos are step-by-step solutions available in
, or on this text’s
Channel. Flip back to the Resources
for Success page for a link to this text’s YouTube channel.
1. Consider the linear function f 1x2 = - 4x + 3:
(a) Find the slope and y-intercept.
(b) What is the average rate of change of f ?
(c) Determine whether f is increasing, decreasing, or constant.
(d) Graph f.
In Problems 2 and 3, find the intercepts, if any, of each quadratic function.
2. f 1x2 = 3x2 - 2x - 8
3. G1x2 = - 2x2 + 4x + 1
4. Given that f 1x2 = x2 + 3x and g1x2 = 5x + 3, solve f 1x2 = g1x2. Graph each function and label the points of intersection.
5. Graph f 1x2 = 1x - 32 2 - 2 using transformations.
6. Consider the quadratic function f 1x2 = 3x2 - 12x + 4:
(a) Determine whether the graph opens up or down.
(b) Determine the vertex.
(c) Determine the axis of symmetry.
(d) Determine the intercepts.
(e) Use the information from parts (a)–(d) to graph f.
7. Determine whether f 1x2 = - 2x2 + 12x + 3 has a maximum or a minimum. Then find the maximum or minimum
value.
8. Solve x2 - 10x + 24 Ú 0.
9. RV Rental The weekly rental cost of a 20-foot recreational vehicle is $129.50 plus $0.15 per mile.
(a) Find a linear function that expresses the cost C as a function of miles driven m.
(b) What is the rental cost if 860 miles are driven?
(c) How many miles were driven if the rental cost is $213.80?
Cumulative Review
319
Cumulative Review
1. Find the distance between the points P = 1 - 1, 32 and
Q = 14, - 22. Find the midpoint of the line segment P to Q.
2. Which of the following points are on the graph of
y = x3 - 3x + 1?
(a) 1 - 2, - 12
(b) (2, 3)
(c) (3, 1)
3. Solve the inequality 5x + 3 Ú 0 and graph the solution set.
4. Find the equation of the line containing the points 1 - 1, 42
and 12, - 22. Express your answer in slope–intercept form
and graph the line.
5. Find the equation of the line perpendicular to the line
y = 2x + 1 and containing the point (3, 5). Express your
answer in slope–intercept form and graph both lines.
6. Graph the equation x2 + y2 - 4x + 8y - 5 = 0.
7. Does the following relation represent a function?
5 1 - 3, 82, 11, 32, 12, 52, 13, 82 6 .
8. For the function f defined by f 1x2 = x2 - 4x + 1, find:
(a) f 122
(b) f 1x2 + f 122
(c) f 1 - x2
(d) - f 1x2
f 1x + h2 - f 1x2
(e) f 1x + 22
(f)
, h ≠ 0
h
3z - 1
.
9. Find the domain of h1z2 =
6z - 7
10. Is the following graph the graph of a function?
x2
even, odd, or neither?
2x + 1
13. Approximate the local maximum values and local minimum
values of f 1x2 = x3 - 5x + 1 on 1 - 4, 42. Determine
where the function is increasing and where it is decreasing.
12. Is the function f 1x2 =
14. If f 1x2 = 3x + 5 and g1x2 = 2x + 1:
(a) Solve f 1x2 = g1x2.
(b) Solve f 1x2 7 g1x2.
15. Consider the graph below of the function f.
(a) Find the domain and the range of f.
(b) Find the intercepts.
(c) Is the graph of f symmetric with respect to the x-axis, the
y-axis, or the origin?
(d) Find f(2).
(e) For what value(s) of x is f 1x2 = 3?
(f) Solve f 1x2 6 0.
(g) Graph y = f 1x2 + 2.
(h) Graph y = f 1 - x2.
(i) Graph y = 2f 1x2.
(j) Is f even, odd, or neither?
(k) Find the interval(s) on which f is increasing.
y
4
(4, 3)
(2, 1)
y
5
x
11. Consider the function f 1x2 =
(4, 3)
x
.
x + 4
1
(a) Is the point a1, b on the graph of f ?
4
(b) If x = - 2, what is f (x)? What point is on the graph
of f ?
(c) If f 1x2 = 2, what is x? What point is on the graph
of f ?
(2, 1)
(1, 0)
4
(1, 0)
(0, 1)
5
x
320
CHAPTER 4 Linear and Quadratic Functions
Chapter Projects
variable and the percentage change in the stock you
chose as the dependent variable. The easiest way to
draw a scatter diagram in Excel is to place the two
columns of data next to each other (for example, have the
percentage change in the S&P500 in column F and the
percentage change in the stock you chose in column G).
Then highlight the data and select the Scatter Diagram
icon under Insert. Comment on the type of relation that
appears to exist between the two variables.
Internet-based Project
I. The Beta of a Stock You want to invest in the stock
market but are not sure which stock to purchase. Information
is the key to making an informed investment decision. One
piece of information that many stock analysts use is the beta
of the stock. Go to Wikipedia (http://en.wikipedia.org/wiki/
Beta_%28finance%29) and research what beta measures
and what it represents.
1.
3. Finding beta. To find beta requires that we find the line
of best fit using least-squares regression. The easiest
approach is to click inside the scatter diagram. Select the
Chart Elements icon ( + ). Check the box for Trendline,
select the arrow to the right, and choose More Options.
Select Linear and check the box for Display Equation on
chart. The line of best fit appears on the scatter diagram.
See below.
Approximating the beta of a stock. Choose a wellknown company such as Google or Coca-Cola. Go
to a website such as Yahoo! Finance (http://finance.
yahoo.com/) and find the weekly closing price of
the company’s stock for the past year. Then find the
closing price of the Standard & Poor’s 500 (S&P500) for
the same time period.
To get the historical prices in Yahoo!
Finance, select Historical Prices from the left
menu. Choose the appropriate time period. Select
Weekly and Get Prices. Finally, select Download to
Spreadsheet. Repeat this for the S&P500, and copy
the data into the same spreadsheet. Finally, rearrange
the data in chronological order. Be sure to expand the
selection to sort all the data. Now, using the adjusted close
price, compute the percentage change in price for each
P1 - Po
week, using the formula % change =
. For
Po
example, if week 1 price is in cell D1 and week 2 price is
D2 - D1
in cell D2, then % change =
. Repeat this for
D1
the S&P500 data.
2. Using Excel to draw a scatter diagram. Treat the
percentage change in the S&P500 as the independent
y
0.08
0.06
0.04
0.02
0.1
0.05
O
0.05
0.1
x
0.02
0.04
0.06
y 0.9046x 0.0024 R 2 0.4887
Series 1
Linear (series 1)
The line of best fit for this data is y = 0.9046x + 0.0024.
You may click on Chart Title or either axis title and
insert the appropriate names. The beta is the slope of the
line of best fit, 0.9046. We interpret this by saying, “If the
S&P500 increases by 1%, then this stock will increase
by 0.9%, on average.” Find the beta of your stock and
provide an interpretation. NOTE: Another way to use
Excel to find the line of best fit requires using the Data
Analysis Tool Pack under add-ins.
The following projects are available on the Instructor’s Resource Center (IRC):
II. Cannons A battery commander uses the weight of a missile, its initial velocity, and the position of its gun to determine where the
missile will travel.
III. First and Second Differences Finite differences provide a numerical method that is used to estimate the graph of an unknown
function.
IV. CBL Experiment Computer simulation is used to study the physical properties of a bouncing ball.
5
Polynomial and
Rational Functions
Day Length
Day length is the length of time each day from the moment the
upper limb of the sun’s disk appears above the horizon during
sunrise to the moment when the upper limb disappears below the
horizon during sunset. The length of a day depends on the day
of the year as well as the latitude of the location. Latitude gives
the location of a point on Earth north or south of the equator.
In the Internet Project at the end of this chapter, we use
information from the chapter to investigate the relation
between day length and latitude for a specific day of the year.
—See the Internet-based Chapter Project I—
A Look Back
In Chapter 3, we began our discussion of functions. We defined domain, range,
and independent and dependent variables, found the value of a function, and
graphed functions. We continued our study of functions by listing the properties
that a function might have, such as being even or odd, and created a library
of functions, naming key functions and listing their properties, including their
graphs.
In Chapter 4, we discussed linear functions and quadratic functions, which
belong to the class of polynomial functions.
A Look Ahead
In this chapter, we look at two general classes of functions, polynomial functions
and rational functions, and examine their properties. Polynomial functions are
arguably the simplest expressions in algebra. For this reason, they are often used
to approximate other, more complicated functions. Rational functions are ratios
of polynomial functions.
Outline
5.1
5.2
5.3
5.4
5.5
5.6
Polynomial Functions and Models
Properties of Rational Functions
The Graph of a Rational Function
Polynomial and Rational Inequalities
The Real Zeros of a Polynomial
Function
Complex Zeros; Fundamental
Theorem of Algebra
Chapter Review
Chapter Test
Cumulative Review
Chapter Projects
321
322
CHAPTER 5 Polynomial and Rational Functions
5.1 Polynomial Functions and Models
PREPARING FOR THIS SECTION Before getting started, review the following:
r Graphing Techniques: Transformations (Section 3.5,
pp. 247–256)
r Intercepts (Section 2.2, pp. 159–160)
r Polynomials (Chapter R, Section R.4, pp. 39–47)
r Using a Graphing Utility to Approximate Local
Maxima and Local Minima (Section 3.3, p. 229)
r Intercepts of a Function (Section 3.2, pp. 215–217)
Now Work the ‘Are You Prepared?’ problems on page 338.
OBJECTIVES 1 Identify Polynomial Functions and Their Degree (p. 322)
2 Graph Polynomial Functions Using Transformations (p. 326)
3 Identify the Real Zeros of a Polynomial Function and Their
Multiplicity (p. 327)
4 Analyze the Graph of a Polynomial Function (p. 334)
5 Build Cubic Models from Data (p. 336)
1 Identify Polynomial Functions and Their Degree
In Chapter 4, we studied the linear function f1x2 = mx + b, which can be written as
f1x2 = a1x + a0
and the quadratic function f1x2 = ax2 + bx + c, a ≠ 0, which can be written as
f1x2 = a2x2 + a1x + a0
a2 ≠ 0
Each of these functions is an example of a polynomial function.
DEFINITION
A polynomial function in one variable is a function of the form
f 1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0
(1)
where an , an - 1 , c, a1 , a0 are constants, called the coefficients of the
polynomial, n Ú 0 is an integer, and x is the variable. If an ≠ 0, it is called the
leading coefficient, and n is the degree of the polynomial.
The domain of a polynomial function is the set of all real numbers.
In Words
A polynomial function is a sum of
monomials.
EX AMPLE 1
The monomials that make up a polynomial are called its terms. If an ≠ 0,
anxn is called the leading term; a0 is called the constant term. If all of the coefficients
are 0, the polynomial is called the zero polynomial, which has no degree.
Polynomials are usually written in standard form, beginning with the nonzero
term of highest degree and continuing with terms in descending order according to
degree. If a power of x is missing, it is because its coefficient is zero.
Polynomial functions are among the simplest expressions in algebra. They are
easy to evaluate: only addition and repeated multiplication are required. Because of
this, they are often used to approximate other, more complicated functions. In this
section, we investigate properties of this important class of functions.
Identifying Polynomial Functions
Determine which of the following are polynomial functions. For those that are, state
the degree; for those that are not, tell why not. Write each polynomial in standard
form, and then identify the leading term and the constant term.
1
(a) p 1x2 = 5x3 - x2 - 9 (b) f1x2 = x + 2 - 3x4 (c) g1x2 = 1x
4
x2 - 2
(d) h 1x2 = 3
(e) G1x2 = 8
(f) H1x2 = - 2x3 1x - 12 2
x - 1
SECTION 5.1 Polynomial Functions and Models
Solution
323
(a) p is a polynomial function of degree 3, and it is already in standard form. The
leading term is 5x3, and the constant term is - 9.
(b) f is a polynomial function of degree 4. Its standard form is f1x2 = - 3x4 + x + 2.
The leading term is - 3x4, and the constant term is 2.
1
(c) g is not a polynomial function because g1x2 = 1x = x2, so the variable x is
1
raised to the power, which is not a nonnegative integer.
2
(d) h is not a polynomial function. It is the ratio of two distinct polynomials, and the
polynomial in the denominator is of positive degree.
(e) G is a nonzero constant polynomial function so it is of degree 0. The polynomial
is in standard form. The leading term and constant term are both 8.
(f) H 1x2 = - 2x3 1x - 12 2 = - 2x3 1x2 - 2x + 12 = - 2x5 + 4x4 - 2x3. So, H is a
polynomial function of degree 5. Because H1x2 = - 2x5 + 4x4 - 2x3, the
leading term is - 2x5. Since no constant term is shown, the constant term is 0. Do
you see a way to find the degree of H without multiplying it out?
r
Now Work
17
PROBLEMS
AND
21
We have already discussed in detail polynomial functions of degrees 0, 1, and 2. See
Table 1 for a summary of the properties of the graphs of these polynomial functions.
Table 1
Degree
Form
Name
Graph
No degree
f(x) = 0
Zero function
The x-axis
0
f(x) = a0 , a0 ≠ 0
Constant function
Horizontal line with y-intercept a0
1
f(x) = a1 x + a0 , a1 ≠ 0
Linear function
Nonvertical, nonhorizontal line with
slope a1 and y-intercept a0
2
f(x) = a2 x2 + a1 x + a0 , a2 ≠ 0
Quadratic function
Parabola: graph opens up if a2 7 0;
graph opens down if a2 6 0
One objective of this section is to analyze the graph of a polynomial function. If
you take a course in calculus, you will learn that the graph of every polynomial function
is both smooth and continuous. By smooth, we mean that the graph contains no sharp
corners or cusps; by continuous, we mean that the graph has no gaps or holes and
can be drawn without lifting your pencil from the paper. See Figures 1(a) and (b).
y
y
Cusp
Corner
Gap
Hole
x
Figure 1
(a) Graph of a polynomial function:
smooth, continuous
x
(b) Cannot be the graph of a
polynomial function
Power Functions
We begin the analysis of the graph of a polynomial function by discussing power
functions, a special kind of polynomial function.
DEFINITION
In Words
A power function is defined by a
single monomial.
A power function of degree n is a monomial function of the form
f 1x2 = axn
where a is a real number, a ≠ 0, and n 7 0 is an integer.
(2)
324
CHAPTER 5 Polynomial and Rational Functions
Examples of power functions are
f1x2 = 3x
f1x2 = - 5x2
degree 1
degree 2
f1x2 = 8x3
f1x2 = - 5x4
degree 3
degree 4
The graph of a power function of degree 1, f1x2 = ax, is a straight line, with
slope a, that passes through the origin. The graph of a power function of degree 2,
f1x2 = ax2, is a parabola, with vertex at the origin, that opens up if a 7 0 and opens
down if a 6 0.
If we know how to graph a power function of the form f1x2 = xn, a compression
or stretch and, perhaps, a reflection about the x-axis will enable us to obtain the graph
of g1x2 = axn. Consequently, we shall concentrate on graphing power functions of
the form f1x2 = xn.
We begin with power functions of even degree of the form f1x2 = xn, n Ú 2
and n even. The domain of f is the set of all real numbers, and the range is the set of
nonnegative real numbers. Such a power function is an even function. (Do you see
why?). Its graph is symmetric with respect to the y-axis. Its graph always contains
the origin and the points 1 - 1, 12 and 11, 12 .
If n = 2, the graph is the familiar parabola y = x2 that opens up, with vertex
at the origin. If n Ú 4, the graph of f1x2 = xn, n even, will be closer to the x-axis
than the parabola y = x2 if - 1 6 x 6 1, x ≠ 0, and farther from the x-axis than
the parabola y = x2 if x 6 - 1 or if x 7 1. Figure 2(a) illustrates this conclusion.
Figure 2(b) shows the graphs of y = x4 and y = x8 for comparison.
f (x ) = x n
n≥4
n even y
4
2
2
(1, 1)
(– 1, 1)
(0, 0)
3
x
(1, 1)
(0, 0)
–3
(a)
Figure 2
y=x4
y
4
(– 1, 1)
–3
y=x8
y = x2
3
x
(b)
Figure 2 shows that as n increases, the graph of f1x2 = xn, n Ú 2 and n even,
tends to flatten out near the origin and is steeper when x is far from 0. For large n, it
may appear that the graph coincides with the x-axis near the origin, but it does not;
the graph actually touches the x-axis only at the origin (see Table 2). Also, for large n,
it may appear that for x 6 - 1 or for x 7 1 the graph is vertical, but it is not; it is
only increasing very rapidly in these intervals. If the graphs were enlarged many
times, these distinctions would be clear.
x = 0.1
x = 0.3
f (x) = x
8
10
-8
0.0000656
0.0039063
f (x) = x
20
10
-20
3.487 # 10
0.000001
f (x) = x
40
10
-40
1.216 # 10-21
Table 2
x = 0.5
-11
9.095 # 10-13
Seeing the Concept
Graph Y1 = x4, Y2 = x8, and Y3 = x12 using the viewing rectangle - 2 … x … 2, - 4 … y … 16. Then
graph each again using the viewing rectangle - 1 … x … 1, 0 … y … 1. See Figure 3. TRACE along one
of the graphs to confirm that for x close to 0 the graph is above the x-axis and that for x 7 0 the graph
is increasing.
SECTION 5.1 Polynomial Functions and Models
16
325
1
Y3 5
Y1 5 x 4
x 12
Y2 5 x 8
Y1 5 x 4
Y2 5 x 8
22
2
21
24
Figure 3
Y3 5 x 12
1
0
(b)
(a)
Properties of Power Functions, f (x) = xn, n Is a Positive Even Integer
y
3
1. f is an even function, so its graph is symmetric with respect to the y-axis.
2. The domain is the set of all real numbers. The range is the set of nonnegative
real numbers.
3. The graph always contains the points 1 - 1, 12 , 10, 02 , and 11, 12 .
4. As the exponent n increases in magnitude, the graph is steeper when
x 6 - 1 or x 7 1; but for x near the origin, the graph tends to flatten out
and lie closer to the x-axis.
y = x3
y=xn
n≥5
n odd
(1, 1)
–3
3
(0, 0)
x
(–1, –1)
–3
Figure 4
y
3
y = x9
y = x5
Now we consider power functions of odd degree of the form f1x2 = xn, n Ú 3
and n odd. The domain and the range of f are the set of real numbers. Such a
power function is an odd function. (Do you see why?). Its graph is symmetric with
respect to the origin. Its graph always contains the origin and the points 1 - 1, - 12
and 11, 12 .
The graph of f 1x2 = xn when n = 3 has been shown several times and is
repeated in Figure 4. If n Ú 5, the graph of f1x2 = xn, n odd, will be closer to the
x-axis than that of y = x3 if - 1 6 x 6 1 and farther from the x-axis than that of
y = x3 if x 6 - 1 or if x 7 1. Figure 4 illustrates this conclusion. Figure 5 shows the
graphs of y = x5 and y = x9 for further comparison.
It appears that each graph coincides with the x-axis near the origin, but it does
not; each graph actually crosses the x-axis at the origin. Also, it appears that as
x increases the graphs become vertical, but they do not; each graph is just increasing
very rapidly.
(1, 1)
Seeing the Concept
–3
(0, 0)
3
x
(–1, –1)
Graph Y1 = x3, Y2 = x7, and Y3 = x11 using the viewing rectangle - 2 … x … 2, - 16 … y … 16. Then
graph each again using the viewing rectangle - 1 … x … 1, - 1 … y … 1. See Figure 6. TRACE along
one of the graphs to confirm that the graph is increasing and crosses the x-axis at the origin.
–3
16
1
Figure 5
Y1 5 x 3
Y2 5 x 7
Y3 5 x 11
Y2 5 x 7
Y1 5 x 3
22
2
216
Figure 6
(a)
21
Y3 5 x 11
1
21
(b)
326
CHAPTER 5 Polynomial and Rational Functions
To summarize:
Properties of Power Functions, f (x) = xn, n Is a Positive Odd Integer
f is an odd function, so its graph is symmetric with respect to the origin.
The domain and the range are the set of all real numbers.
The graph always contains the points 1 - 1, - 12 , 10, 02 , and 11, 12 .
As the exponent n increases in magnitude, the graph is steeper when
x 6 - 1 or x 7 1; but for x near the origin, the graph tends to flatten out
and lie closer to the x-axis.
1.
2.
3.
4.
2 Graph Polynomial Functions Using Transformations
The methods of shifting, compression, stretching, and reflection (studied in Section 3.5),
when used with the facts just presented, will enable us to graph polynomial functions
that are transformations of power functions.
EX AMPLE 2
Graphing a Polynomial Function Using Transformations
Graph: f1x2 = 1 - x5
Solution
It is helpful to rewrite f as f1x2 = - x5 + 1. Figure 7 shows the required stages.
y
y
2
2
(1, 1)
2
(0, 0)
x
2
2
(1, 0)
(0, 0)
x
2
x
2
2
2
Multiply by 1;
reflect
about x-axis
Add 1;
shift up
1 unit
(b) y x 5
(a) y x 5
(c) y x 5 1 1 x 5
r
Graphing a Polynomial Function Using Transformations
Graph: f1x2 =
Solution
2
(1, 1)
2
EX AMPLE 3
(0, 1)
(1, 1)
(1, 1)
Figure 7
y
2
(1, 2)
1
1x - 12 4
2
Figure 8 shows the required stages.
y
2
y
2
(1, 1)
(1, 1)
2
(0, 0)
2
y
2
(0, 1)
x
2
2
(2, 1)
(1, 0) 2
x
(0, 1–2)
(2, 1–2)
2
(1, 0) 2
x
2
2
1–
;
2
Figure 8
(a) y x 4
Multiply by
Replace x by x 1;
shift right
compression by
1 unit
a factor of 1–2
(c) y (b) y ( x 1)4
Now Work
PROBLEMS
29
AND
35
1–
2
( x 1)4
r
SECTION 5.1 Polynomial Functions and Models
327
3 Identify the Real Zeros of a Polynomial Function
and Their Multiplicity
Figure 9 shows the graph of a polynomial function with four x-intercepts. Notice
that at the x-intercepts, the graph must either cross the x-axis or touch the x-axis.
Consequently, between consecutive x-intercepts the graph is either above the x-axis
or below the x-axis.
y
Above
x-axis
Above
x -axis
x
Crosses
x -axis
Below
x -axis
Touches
x -axis
Crosses
x -axis
Below
x -axis
Figure 9 Graph of a polynomial function
If a polynomial function f is factored completely, it is easy to locate the
x-intercepts of the graph by solving the equation f1x2 = 0 and using
the Zero-Product Property. For example, if f 1x2 = 1x - 12 2 1x + 32 , then the
solutions of the equation
f1x2 = 1x - 12 2 1x + 32 = 0
are identified as 1 and - 3. That is, f112 = 0 and f 1 - 32 = 0.
DEFINITION
If f is a function and r is a real number for which f 1r2 = 0, then r is called a
real zero of f.
As a consequence of this definition, the following statements are equivalent.
1.
2.
3.
4.
r is a real zero of a polynomial function f.
r is an x-intercept of the graph of f.
x - r is a factor of f.
r is a solution to the equation f1x2 = 0.
So the real zeros of a polynomial function are the x-intercepts of its graph, and
they are found by solving the equation f1x2 = 0.
EXAMPL E 4
Finding a Polynomial Function from Its Zeros
(a) Find a polynomial function of degree 3 whose zeros are - 3, 2, and 5.
(b) Use a graphing utility to graph the polynomial found in part (a) to verify your
result.
Solution
(a) If r is a real zero of a polynomial function f, then x - r is a factor of f. This means
that x - 1 - 32 = x + 3, x - 2, and x - 5 are factors of f. As a result, any
polynomial function of the form
f1x2 = a1x + 32 1x - 22 1x - 52
where a is a nonzero real number, qualifies. The value of a causes a stretch,
compression, or reflection, but it does not affect the x-intercepts of the graph.
Do you know why?
328
CHAPTER 5 Polynomial and Rational Functions
(b) We choose to graph f with a = 1. Then
40
24
f1x2 = 1x + 32 1x - 22 1x - 52 = x3 - 4x2 - 11x + 30
6
Figure 10 shows the graph of f. Notice that the x-intercepts are - 3, 2, and 5.
r
Seeing the Concept
250
Figure 10 f(x) = x3 - 4x2 - 11x + 30
Graph the function found in Example 4 for a = 2 and a = - 1. Does the value of a affect the zeros of f?
How does the value of a affect the graph of f?
Now Work
PROBLEM
43
If the same factor x - r occurs more than once, r is called a repeated, or
multiple, zero of f. More precisely, we have the following definition.
DEFINITION
EX AMPLE 5
If 1x - r2 m is a factor of a polynomial f and 1x - r2 m + 1 is not a factor of f,
then r is called a zero of multiplicity m of f.*
Identifying Zeros and Their Multiplicities
For the polynomial
f1x2 = 5x2 1x + 22 ax -
In Words
The multiplicity of a zero is the
number of times its corresponding
factor occurs.
1 4
b
2
r - 2 is a zero of multiplicity 1 because the exponent on the factor x + 2 is 1.
r 0 is a zero of multiplicity 2 because the exponent on the factor x is 2.
1
1
r
is a zero of multiplicity 4 because the exponent on the factor x - is 4.
2
2
Now Work
PROBLEM
r
57(a)
Suppose that it is possible to completely factor a polynomial function and, as a
result, locate all the x-intercepts of its graph (the real zeros of the function). These
x-intercepts then divide the x-axis into open intervals and, on each such interval,
the graph of the polynomial will be either above or below the x-axis over the entire
interval. Let’s look at an example.
EX AMPLE 6
Graphing a Polynomial Using Its x-Intercepts
Consider the following polynomial: f 1x2 = 1x + 12 2 1x - 22
(a) Find the x- and y-intercepts of the graph of f.
(b) Use the x-intercepts to find the intervals on which the graph of f is above the
x-axis and the intervals on which the graph of f is below the x-axis.
(c) Locate other points on the graph, and connect all the points plotted with a
smooth, continuous curve.
Solution
(a) The y-intercept is f 102 = 10 + 12 2 10 - 22 = - 2.
The x-intercepts satisfy the equation
f1x2 = 1x + 12 2 1x - 22 = 0
from which we find
1x + 12 2 = 0
or x - 2 = 0
x = - 1 or
x = 2
The x-intercepts are - 1 and 2.
*Some books use the terms multiple root and root of multiplicity m.
SECTION 5.1 Polynomial Functions and Models
329
(b) The two x-intercepts divide the x-axis into three intervals:
1 - q , - 12
1 - 1, 22
12, q 2
Since the graph of f crosses or touches the x-axis only at x = - 1 and x = 2, it
follows that the graph of f is either above the x-axis 3 f 1x2 7 04 or below the
x-axis 3 f1x2 6 04 on each of these three intervals. To see where the graph lies,
we need only pick a number in each interval, evaluate f there, and see whether the
value is positive (above the x-axis) or negative (below the x-axis). See Table 3.
(c) In constructing Table 3, we obtained three additional points on the graph:
1 - 2, - 42, 11, - 42 and 13, 162 . Figure 11 illustrates these points, the intercepts,
and a smooth, continuous curve (the graph of f ) connecting them.
y
–1
Table 3
2
x
(3, 16)
Interval
( - q , - 1)
( - 1, 2)
(2, q )
Number chosen
-2
1
3
Value of f
f ( - 2) = - 4
f (1) = - 4
f (3) = 16
Location of graph
Below x-axis
Below x-axis
Above x-axis
Point on graph
( - 2, - 4)
(1, - 4)
(3, 16)
12
6
(– 1, 0)
(2, 0)
3
–2
(– 2, – 4)
x
(0, – 2)
–6
(1, –4)
Figure 11 f 1x2 = 1x + 12 2 1x - 22
r
Look again at Table 3. Since the graph of f1x2 = 1x + 12 2 1x - 22 is below
the x-axis on both sides of - 1, the graph of f touches the x-axis at x = - 1, a zero of
multiplicity 2. Since the graph of f is below the x-axis for x 6 2 and above the x-axis
for x 7 2, the graph of f crosses the x-axis at x = 2, a zero of multiplicity 1.
This suggests the following results:
If r Is a Zero of Even Multiplicity
Numerically: The sign of f 1x2 does not change from one side to the other side of r.
Graphically: The graph of f touches the x-axis at r.
If r Is a Zero of Odd Multiplicity
Numerically: The sign of f1x2 changes from one side to the other side of r.
Graphically: The graph of f crosses the x-axis at r.
Now Work
PROBLEM
57(b)
Turning Points
Look again at Figure 11 above. We cannot be sure just how low the graph actually
goes between x = - 1 and x = 2. But we do know that somewhere in the interval
1 - 1, 22 the graph of f must change direction (from decreasing to increasing). The
points at which a graph changes direction are called turning points.* Each turning
point yields either a local maximum or a local minimum (see Section 3.3). The
following result from calculus tells us the maximum number of turning points that
the graph of a polynomial function can have.
*Graphing utilities can be used to approximate turning points. For most polynomials, calculus is needed
to find the exact turning points.
330
CHAPTER 5 Polynomial and Rational Functions
THEOREM
Turning Points
If f is a polynomial function of degree n, then the graph of f has at most n - 1
turning points.
If the graph of a polynomial function f has n - 1 turning points, then the
degree of f is at least n.
Based on the first part of the theorem, a polynomial function of degree 5 will
have at most 5 - 1 = 4 turning points. Based on the second part of the theorem, if
the graph of a polynomial function has three turning points, then the degree of the
function must be at least 4.
Exploration
A graphing utility can be used to locate the turning points of a graph. Graph Y1 = (x + 1)2(x - 2). Use
MINIMUM to find the location of the turning point for 0 6 x 6 2. See Figure 12.
0
2
26
Figure 12 Y1 = 1x + 12 2 1x - 22
Now Work
EX AMPLE 7
PROBLEM
57(c)
Identifying the Graph of a Polynomial Function
Which of the graphs in Figure 13 could be the graph of a polynomial function? For
those that could, list the real zeros and state the least degree the polynomial can
have. For those that could not, say why not.
y
y
y
2
2
2
–2
2
x
–2
–2
2
–2
x
–2
y
3
2
x
3 x
–3
–2
–3
Figure 13
Solution
(a)
(b)
(c)
(d)
(a) The graph in Figure 13(a) cannot be the graph of a polynomial function because
of the gap that occurs at x = - 1. Remember, the graph of a polynomial function
is continuous—no gaps or holes. (See Figure 1.)
(b) The graph in Figure 13(b) could be the graph of a polynomial function because the
graph is smooth and continuous. It has three real zeros: - 2, 1, and 2. Since the graph
has two turning points, the degree of the polynomial function must be at least 3.
(c) The graph in Figure 13(c) cannot be the graph of a polynomial function because
of the cusp at x = 1. Remember, the graph of a polynomial function is smooth.
(d) The graph in Figure 13(d) could be the graph of a polynomial function. It has
two real zeros: - 2 and 1. Since the graph has three turning points, the degree of
the polynomial function is at least 4.
r
Now Work
PROBLEM
69
SECTION 5.1 Polynomial Functions and Models
331
End Behavior
One last remark about Figure 11. For very large values of x, either positive or negative,
the graph of f1x2 = 1x + 12 2 1x - 22 looks like the graph of y = x3. To see why,
we write f in the form
f 1x2 = 1x + 12 2 1x - 22 = x3 - 3x - 2 = x3 a1 -
3
2
- 3b
2
x
x
3
2
Now, for large values of x, either positive or negative, the terms 2 and 3 are
x
x
close to 0, so for large values of x,
f1x2 = x3 - 3x - 2 = x3 a1 -
3
2
- 3 b ≈ x3
x2
x
The behavior of the graph of a function for large values of x, either positive or
negative, is referred to as its end behavior.
THEOREM
End Behavior
For large values of x, either positive or negative, the graph of the polynomial
function
f1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0
In Words
The end behavior of a polynomial
function resembles that of its
leading term.
an ≠ 0
resembles the graph of the power function
y = an xn
For example, if f 1x2 = - 2x3 + 5x2 + x - 4, then the graph of f will behave
like the graph of y = - 2x3 for very large values of x, either positive or negative.
We can see that the graphs of f and y = - 2x3 “behave” the same by considering
Table 4 and Figure 14.
Table 4
x
f(x)
y = − 2x3
10
- 1,494
- 2,000
100
- 1,949,904
- 2,000,000
500
- 248,749,504
- 250,000,000
1,000
- 1,994,999,004
- 2,000,000,000
175
Y2 5 22x 3 1 5x 2 1 x 2 4
25
5
Y1 5 22x 3
2175
Figure 14
NOTE Infinity ( q ) and negative infinity
( - q ) are not numbers. Rather, they are
symbols that represent unboundedness.
■
Notice that as x becomes a larger and larger positive number, the values of f
become larger and larger negative numbers. When this happens, we say that f is
unbounded in the negative direction. Rather than using words to describe the
behavior of the graph of the function, we explain its behavior using notation. We
can symbolize “the value of f becomes a larger and larger negative number as x
becomes a larger and larger positive number” by writing f1x2 S - q as x S q
(read “the values of f approach negative infinity as x approaches infinity”).
In calculus, limits are used to convey these ideas. There we use the symbolism
lim
f1x2 = - q , read “the limit of f1x2 as x approaches infinity equals negative
xS q
infinity,” to mean that f1x2 S - q as x S q .
332
CHAPTER 5 Polynomial and Rational Functions
When we say that the value of a limit equals infinity (or negative infinity), we
mean that the values of the function are unbounded in the positive (or negative)
direction and call the limit an infinite limit. When we discuss limits as x becomes
unbounded in the negative direction or unbounded in the positive direction, we are
discussing limits at infinity.
Look back at Figures 2 and 4. Based on the preceding theorem and the previous
discussion on power functions, the end behavior of a polynomial function can be of
only four types. See Figure 15.
f(x)
as x
`
2`
y
f(x)
as x
`
`
y
y
x
f(x)
as x
`
`
x
2`
2`
f(x)
as x
f(x)
as x
E
n ≥ 2 even; an < 0
D
n ≥ 2 even; an > 0
2`
`
f(x)
as x
`
2` y
x
f(x)
as x
x
2`
2`
f(x)
as x
F
n ≥ 3 odd; an > 0
G
n ≥ 3 odd; an < 0
2`
`
Figure 15 End behavior of f(x) = anxn + an-1xn-1 + g + a1x + a0
For example, if f1x2 = - 2x4 + x3 + 4x2 - 7x + 1, the graph of f will resemble
the graph of the power function y = - 2x4 for large 0 x 0 . The graph of f will behave
like Figure 15(b) for large 0 x 0 .
Now Work
EX AMPLE 8
PROBLEM
57(d)
Identifying the Graph of a Polynomial Function
Which of the graphs in Figure 16 could be the graph of
f(x) = x4 + ax3 + bx2 - 5x - 6
where a 7 0, b 7 0?
y
y
x
(a)
y
x
(b)
y
x
(c)
x
(d)
Figure 16
Solution
The y-intercept of f is f 102 = - 6. We can eliminate the graph in Figure 16(a), whose
y-intercept is positive.
We are not able to solve f1x2 = 0 to find the x-intercepts of f, so we move on to
investigate the turning points of each graph. Since f is of degree 4, the graph of f has
at most 3 turning points. We eliminate the graph in Figure 16(c) because that graph
has 5 turning points.
Now we look at end behavior. For large values of x, the graph of f will behave
like the graph of y = x4. This eliminates the graph in Figure 16(d), whose end behavior
is like the graph of y = - x4.
SECTION 5.1 Polynomial Functions and Models
333
Only the graph in Figure 16(b) could be the graph of
f 1x2 = x4 + ax3 + bx2 - 5x - 6
where a 7 0, b 7 0.
EXAMPL E 9
r
Writing a Polynomial Function from Its Graph
Write a polynomial function whose graph is shown in Figure 17 (use the smallest
degree possible).
y
15
(– 1, 6)
– 3 (– 2, 0)
9
(0, 0)
(2, 0)
3 x
–9
– 15
– 21
Figure 17
Solution
The x-intercepts are - 2, 0, and 2. Therefore, the polynomial must have the factors
1x + 22 , x, and 1x - 22, respectively. There are three turning points, so the degree
of the polynomial must be at least 4. The graph touches the x-axis at x = - 2, so - 2
must have an even multiplicity. The graph crosses the x-axis at x = 0 and x = 2, so
0 and 2 must have odd multiplicities. Using the smallest degree possible (1 for odd
multiplicity and 2 for even multiplicity), we can write
f1x2 = ax1x + 2221x - 22
All that remains is to find the leading coefficient, a. From Figure 17, the point 1 - 1, 62
must lie on the graph.
6 = a1 - 12 1 - 1 + 2221 - 1 - 22
6 = 3a
2 = a
f (- 1) = 6
The polynomial function f1x2 = 2x1x + 2221x - 22 would have the graph in
Figure 17.
Check: Graph Y1 = 2x1x + 22 2 1x - 22 using a graphing utility to verify this
result.
r
Now Work
PROBLEMS
73
AND
77
SUMMARY
Graph of a Polynomial Function f 1 x2 = an xn + an - 1 xn - 1 + g + a1x + a0 an 3 0
Degree of the polynomial function f : n
y-intercept: f102 = a0.
Graph is smooth and continuous.
Maximum number of turning points: n - 1
At a zero of even multiplicity: The graph of f touches the x-axis.
At a zero of odd multiplicity: The graph of f crosses the x-axis.
Between zeros, the graph of f is either above or below the x-axis.
End behavior: For large 0 x 0 , the graph of f behaves like the graph of y = an xn.
334
CHAPTER 5 Polynomial and Rational Functions
4 Analyze the Graph of a Polynomial Function
How to Analyze the Graph of a Polynomial Function
EX A MPL E 1 0
Analyze the factored form of the polynomial function f1x2 = 12x + 12 1x - 32 2.
Step-by-Step Solution
Expand the polynomial:
f1x2 = 12x + 12 1x - 32 2 = 12x + 12 1x2 - 6x + 92
Step 1: Determine the end behavior
of the graph of the function.
= 2x3 - 12x2 + 18x + x2 - 6x + 9
Multiply.
= 2x - 11x + 12x + 9
Combine like terms.
3
2
The polynomial function f is of degree 3. The graph of f behaves like y = 2x3 for
large values of x .
The y-intercept is f102 = 9. To find the x-intercepts, solve f 1x2 = 0.
Step 2: Find the x- and y-intercepts
of the graph of the function.
f1x2 = 0
12x + 12 1x - 32 2 = 0
2x + 1 = 0
x = The x-intercepts are -
1
2
or
1x - 32 2 = 0
or
x = 3
1
and 3.
2
Step 3: Determine the zeros of the
function and their multiplicity. Use
this information to determine whether
the graph crosses or touches the
x-axis at each x-intercept.
1
1
and 3. The zero - is a zero of multiplicity 1, so the graph of f
2
2
1
crosses the x-axis at x = - . The zero 3 is a zero of multiplicity 2, so the graph of f
2
touches the x-axis at x = 3.
Step 4: Determine the maximum
number of turning points on the
graph of the function.
Because the polynomial function is of degree 3 (Step 1), the graph of the function
will have at most 3 - 1 = 2 turning points.
Step 5: Put all the information from
Steps 1 through 4 together to obtain
the graph of f. To help establish the
y-axis scale, find additional points
on the graph on each side of any
x-intercept.
Figure 18(a) illustrates the information obtained from Steps 1 through 4. We
evaluate f at - 1, 1, and 4 to help establish the scale on the y-axis.
We find that f1 - 12 = - 16, f112 = 12, and f142 = 9, so we plot the points
1 - 1, - 162, 11, 122 , and (4, 9). The graph of f is given in Figure 18(b).
The zeros of f are -
y
y
End behavior:
Resembles
y = 2x 3
40
40
30
The graph
y -intercept: 9
20
crosses the
x-axis at
1
− 1– , 0 10
x = − –2 . ( 2 )
(3, 0)
(0, 9)
−2
−1
−10
1
Figure 18
3
20
(
− 1–2 ,
4
The graph
touches the
x-axis at
x = 3.
−20
End behavior:
Resembles
y = 2x 3
2
−30
−40
30
5
x
0) 10
−2
−1
−10
(−1, −16)
−20
(1, 12)
1
(4, 9)
(3, 0)
(0, 9)
2
3
4
5
x
−30
−40
(a)
(b)
r
SECTION 5.1 Polynomial Functions and Models
335
SUMMARY
Analyzing the Graph of a Polynomial Function
STEP 1: Determine the end behavior of the graph of the function.
STEP 2: Find the x- and y-intercepts of the graph of the function.
STEP 3: Determine the zeros of the function and their multiplicity. Use this information to determine whether the
graph crosses or touches the x-axis at each x-intercept.
STEP 4: Determine the maximum number of turning points on the graph of the function.
STEP 5: Use the information in Steps 1 through 4 to draw a complete graph of the function. To help establish the
y-axis scale, find additional points on the graph on each side of any x-intercept.
Now Work
PROBLEM
81
For polynomial functions that have noninteger coefficients and for polynomials
that are not easily factored, we use a graphing utility early in the analysis. This
is because the amount of information that can be obtained from algebraic analysis
is limited.
EX AM PL E 1 1
How to Use a Graphing Utility to Analyze the Graph
of a Polynomial Function
Analyze the graph of the polynomial function
f1x2 = x3 + 2.48x2 - 4.3155x + 2.484406
Step-by-Step Solution
Step 1: Determine the end behavior
of the graph of the function.
The polynomial function f is of degree 3. The graph of f behaves like y = x3 for large
values of x .
Step 2: Graph the function using a
graphing utility.
See Figure 19 for the graph of f.
15
25
3
210
Figure 19 f(y1) = x3 + 2.48x2 - 4.3155x + 2.484406
Step 3: Use a graphing utility to
approximate the x- and y-intercepts
of the graph.
The y-intercept is f102 = 2.484406.
In Example 10, the polynomial function was factored, so it was easy to find the
x-intercepts algebraically. However, it is not readily apparent how to factor f in
this example. Therefore, we use a graphing utility’s ZERO (or ROOT or SOLVE)
feature and find the lone x-intercept to be - 3.79, rounded to two decimal places.
336
CHAPTER 5 Polynomial and Rational Functions
Step 4: Use a graphing utility to
create a TABLE to find points on the
graph around each x-intercept.
Table 5 below shows values of x on each side of the x-intercept. The points
1 - 4, - 4.572 and 1 - 2, 13.042 are on the graph.
Table 5
Step 5: Approximate the turning
points of the graph.
From the graph of f shown in Figure 19, we can see that f has two turning points.
Using MAXIMUM reveals one turning point is at ( - 2.28, 13.36), rounded to two
decimal places. Using MINIMUM shows that the other turning point is at (0.63, 1),
rounded to two decimal places.
Step 6: Use the information in
Steps 1 through 5 to draw a complete
graph of the function by hand.
Figure 20 shows a graph of f drawn by hand using the information in Steps 1 through 5.
y
(–2.28, 13.36)
12
(–2, 13.04)
End behavior:
Resembles y = x 3
6
(0, 2.484406)
–5
Figure 20
Step 7: Find the domain and the
range of the function.
Step 8: Use the graph to determine
where the function is increasing and
where it is decreasing.
End behavior:
Resembles y = x 3
(–3.79, 0)
(–4, –4.57)
–6
2x
(0.63, 1)
The domain and the range of f are the set of all real numbers.
Based on the graph, f is increasing on the intervals 1 - q , - 2.282 and 10.63, q 2 .
Also, f is decreasing on the interval 1 - 2.28, 0.632 .
r
SUMMARY
Using a Graphing Utility to Analyze the Graph of a Polynomial Function
STEP 1:
STEP 2:
STEP 3:
STEP 4:
STEP 5:
STEP 6:
STEP 7:
STEP 8:
Determine the end behavior of the graph of the function.
Graph the function using a graphing utility.
Use a graphing utility to approximate the x- and y-intercepts of the graph.
Use a graphing utility to create a TABLE to find points on the graph around each x-intercept.
Approximate the turning points of the graph.
Use the information in Steps 1 through 5 to draw a complete graph of the function by hand.
Find the domain and the range of the function.
Use the graph to determine where the function is increasing and where it is decreasing.
Now Work
PROBLEM
99
5 Build Cubic Models from Data
In Section 4.2 we found the line of best fit from data, and in Section 4.4 we found
the quadratic function of best fit. It is also possible to find polynomial functions of
SECTION 5.1 Polynomial Functions and Models
337
best fit. However, most statisticians do not recommend finding polynomials of best
fit of degree higher than 3.
Data that follow a cubic relation should look like Figure 21(a) or (b).
y 5 ax 3 1 bx 2 1 cx 1 d, a . 0
y 5 ax 3 1 bx 2 1 cx 1 d, a , 0
D
E
Figure 21 Cubic relation
A Cubic Function of Best Fit
EX AM PL E 1 2
The data in Table 6 represent the weekly cost C (in thousands of dollars) of printing
x thousand textbooks.
Table 6
Number x of
Textbooks,
(thousands)
(a) Draw a scatter diagram of the data using x as the independent variable and
C as the dependent variable. Comment on the type of relation that may exist
between the two variables x and C.
(b) Using a graphing utility, find the cubic function of best fit C = C 1x2 that
models the relation between number of texts and cost.
(c) Graph the cubic function of best fit on your scatter diagram.
(d) Use the function found in part (b) to predict the cost of printing 22 thousand
texts per week.
Cost, C
($1000s)
0
100
5
128.1
10
144
13
153.5
17
161.2
Solution
18
162.6
20
166.3
23
178.9
25
190.2
27
221.8
(a) Figure 22 shows the scatter diagram. A cubic relation may exist between the two
variables.
(b) Upon executing the CUBIC REGression program, we obtain the results
shown in Figure 23. The output that the utility provides shows us the
equation y = ax3 + bx2 + cx + d. The cubic function of best fit to the data is
C 1x2 = 0.0155x3 - 0.5951x2 + 9.1502x + 98.4327.
(c) Figure 24 shows the graph of the cubic function of best fit on the scatter diagram.
The function fits the data reasonably well.
250
22
250
22
30
0
Figure 22
Figure 23
0
30
Figure 24
(d) Evaluate the function C 1x2 at x = 22.
C 1222 = 0.01551222 3 - 0.59511222 2 + 9.15021222 + 98.4327 ≈ 176.8
The model predicts that the cost of printing 22 thousand textbooks in a week
will be 176.8 thousand dollars—that is, $176,800.
r
338
CHAPTER 5 Polynomial and Rational Functions
5.1 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. The intercepts of the equation 9x2 + 4y = 36 are
(pp. 159–160)
.
2. Is the expression 4x3 - 3.6x2 - 12 a polynomial? If so,
what is its degree? (pp. 39–47)
3. To graph y = x - 4, you would shift the graph of y = x
a distance of
units. (pp. 247–256)
2
2
5. True or False The x-intercepts of the graph of a function
y = f 1x2 are the real solutions of the equation f 1x2 = 0.
(pp. 215–217)
6. If g152 = 0, what point is on the graph of g? What is the
corresponding x-intercept of the graph of g? (pp. 215–217)
4. Use a graphing utility to approximate (rounded to two decimal
places) the local maximum value and local minimum value of
f 1x2 = x3 - 2x2 - 4x + 5, for - 3 6 x 6 3. (p. 229)
Concepts and Vocabulary
12. The graph of the function f 1x2 = 3x4 - x3 + 5x2 - 2x - 7
will behave like the graph of
for large values of x .
7. The graph of every polynomial function is both
and
.
8. If r is a real zero of even multiplicity of a polynomial function f,
(crosses/touches) the x-axis at r.
then the graph of f
9. The graphs of power functions of the form f 1x2 = xn, where
n is an even integer, always contain the points
,
, and
.
10. If r is a solution to the equation f 1x2 = 0, name three
additional statements that can be made about f and r,
assuming f is a polynomial function.
11. The points at which a graph changes direction (from
increasing to decreasing or decreasing to increasing) are
called
.
13. If f 1x2 = - 2x5 + x3 - 5x2 + 7, then
and lim f 1x2 =
.
lim f 1x2 =
xS -q
x Sq
14. Explain what the notation lim f 1x2 = - q means.
x Sq
of a zero is the number of times its corresponding
15. The
factor occurs.
(a) degree (b) multiplicity (c) turning point (d) limit
16. The graph of y = 5x6 - 3x4 + 2x - 9 has at most how many
turning points?
(a) - 9 (b) 14 (c) 6 (d) 5
Skill Building
In Problems 17–28, determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell
why not. Write each polynomial in standard form. Then identify the leading term and the constant term.
1 - x2
17. f 1x2 = 4x + x3
18. f 1x2 = 5x2 + 4x4
19. g1x2 =
2
20. h1x2 = 3 -
1
x
2
23. g1x2 = x3/2 - x2 + 2
26. F 1x2 =
x2 - 5
x3
21. f 1x2 = 1 -
22. f 1x2 = x1x - 12
1
x
24. h1x2 = 1x1 1x - 12
25. F 1x2 = 5x4 - px3 +
27. G1x2 = 21x - 12 2 1x2 + 12
28. G1x2 = - 3x2 1x + 22 3
1
2
In Problems 29–42, use transformations of the graph of y = x4 or y = x5 to graph each function.
29. f 1x2 = 1x + 12 4
30. f 1x2 = 1x - 22 5
31. f 1x2 = x5 - 3
32. f 1x2 = x4 + 2
33. f 1x2 =
34. f 1x2 = 3x5
35. f 1x2 = - x5
36. f 1x2 = - x4
37. f 1x2 = 1x - 12 5 + 2
38. f 1x2 = 1x + 22 4 - 3
39. f 1x2 = 21x + 12 4 + 1
40. f 1x2 =
41. f 1x2 = 4 - 1x - 22 5
42. f 1x2 = 3 - 1x + 22 4
1 4
x
2
1
1x - 12 5 - 2
2
339
SECTION 5.1 Polynomial Functions and Models
In Problems 43–50, form a polynomial function whose real zeros and degree are given. Answers will vary depending on the choice of a
leading coefficient.
43. Zeros: - 1, 1, 3; degree 3
44. Zeros: - 2, 2, 3; degree 3
45. Zeros: - 3, 0, 4; degree 3
46. Zeros: - 4, 0, 2; degree 3
47. Zeros: - 4, - 1, 2, 3; degree 4
48. Zeros: - 3, - 1, 2, 5; degree 4
49. Zeros: - 1, multiplicity 1; 3, multiplicity 2; degree 3
50. Zeros: - 2, multiplicity 2; 4, multiplicity 1; degree 3
In Problems 51–56, find the polynomial function with the given zeros whose graph passes through the given point.
51. Zeros: - 3, 1, 4
Point: 16, 1802
52. Zeros: - 2, 0, 2
Point: 1 - 4, 162
53. Zeros: - 1, 0, 2, 4
54. Zeros: - 5, - 1, 2, 6
55. Zeros: - 1 (multiplicity 2),
1 (multiplicity 2)
Point: 1 - 2, 452
56. Zeros: - 1 (multiplicity 2),
0, 3 (multiplicity 2)
Point: 11,- 482
5
Point: ¢ , 15≤
2
1
Point: ¢ , 63≤
2
In Problems 57–68, for each polynomial function:
(a) List each real zero and its multiplicity.
(b) Determine whether the graph crosses or touches the x-axis at each x-intercept.
(c) Determine the maximum number of turning points on the graph.
(d) Determine the end behavior; that is, find the power function that the graph of f resembles for large values of x .
57. f 1x2 = 31x - 72 1x + 32 2
58. f 1x2 = 41x + 42 1x + 32 3
60. f 1x2 = 21x - 32 1x2 + 42 3
61. f 1x2 = - 2ax +
63. f 1x2 = 1x - 52 3 1x + 42 2
64. f 1x2 = 1x + 132 2 1x - 22 4
65. f 1x2 = 31x2 + 82 1x2 + 92 2
66. f 1x2 = - 21x2 + 32 3
67. f 1x2 = - 2x2 1x2 - 22
68. f 1x2 = 4x1x2 - 32
59. f 1x2 = 41x2 + 12 1x - 22 3
1 2
b 1x + 42 3
2
62. f 1x2 = ax -
1 2
b 1x - 12 3
3
In Problems 69–72, identify which of the graphs could be the graph of a polynomial function. For those that could, list the real zeros
and state the least degree the polynomial can have. For those that could not, say why not.
69.
70.
y
4
71.
y
4
72.
y
y
4
2
2
2
–2
–4
4 x
2
–2
–4
2
–2
–2
–2
–4
–4
2
4 x
2
x
–2
4
2
2
4
x
2
In Problems 73–76, construct a polynomial function that might have the given graph. (More than one answer may be possible.)
73.
74. y
y
0
1
2
75.
x
0
1
2
76.
y
y
2
2
1
1
x
–2
–1
1
2
3
x
–2
–1
1
–1
–1
–2
–2
2
3
x
340
CHAPTER 5 Polynomial and Rational Functions
In Problems 77–80, write a polynomial function whose graph is shown (use the smallest degree possible).
77.
78.
y
10
79.
y
14
80.
y
72
(3, 8)
(– 2, 16)
6x
–6
6 x
–6
6x
–6
–2
x
2
(2, – 50)
(1, –8)
–14
y
21
– 10
– 15
– 72
In Problems 81–98, analyze each polynomial function by following Steps 1 through 5 on page 335.
81. f 1x2 = x2 1x - 32
82. f 1x2 = x1x + 22 2
83. f 1x2 = 1x + 42 2 11 - x2
84. f 1x2 = 1x - 12 1x + 32 2
85. f 1x2 = - 21x + 22 1x - 22 3
1
86. f 1x2 = - 1x + 42 1x - 12 3
2
87. f 1x2 = 1x + 12 1x - 22 1x + 42
88. f 1x2 = 1x - 12 1x + 42 1x - 32
89. f 1x2 = x2 1x - 22 1x + 22
90. f 1x2 = x2 1x - 32 1x + 42
91. f 1x2 = 1x + 12 2 1x - 22 2
92. f 1x2 = 1x - 42 2 1x + 22 2
93. f 1x2 = x2 1x + 32 1x + 12
94. f 1x2 = x2 1x - 32 1x - 12
95. f 1x2 = 5x1x2 - 42 1x + 32
96. f 1x2 = 1x - 22 2 1x + 22 1x + 42
97. f 1x2 = x2 1x - 22 1x2 + 32
98. f 1x2 = x2 1x2 + 12 1x + 42
In Problems 99–106, analyze each polynomial function f by following Steps 1 through 8 on page 336.
99. f 1x2 = x3 + 0.2x2 - 1.5876x - 0.31752
100. f 1x2 = x3 - 0.8x2 - 4.6656x + 3.73248
101. f 1x2 = x3 + 2.56x2 - 3.31x + 0.89
102. f 1x2 = x3 - 2.91x2 - 7.668x - 3.8151
103. f 1x2 = x4 - 2.5x2 + 0.5625
104. f1x2 = x4 - 18.5x2 + 50.2619
105. f 1x2 = 2x4 - px3 + 15x - 4
106. f 1x2 = - 1.2x4 + 0.5x2 - 13x + 2
Mixed Practice
In Problems 107–114, analyze each polynomial function by following Steps 1 through 5 on page 335.
[Hint: You will need to first factor the polynomial].
107. f 1x2 = 4x - x3
108. f 1x2 = x - x3
109. f 1x2 = x3 + x2 - 12x
110. f 1x2 = x3 + 2x2 - 8x
111. f 1x2 = 2x4 + 12x3 - 8x2 - 48x
112. f1x2 = 4x3 + 10x2 - 4x - 10
113. f 1x2 = - x5 - x4 + x3 + x2
114. f 1x2 = - x5 + 5x4 + 4x3 - 20x2
In Problems 115–118, construct a polynomial function f with the given characteristics.
116. Zeros: - 4, - 1, 2; degree 3; y-intercept: 16
115. Zeros: - 3, 1, 4; degree 3; y-intercept: 36
117. Zeros: - 5(multiplicity 2); 2 (multiplicity 1); 4 (multiplicity 1);
degree 4; contains the point (3, 128)
118. Zeros: - 4 (multiplicity 1); 0 (multiplicity 3); 2 (multiplicity 1);
degree 5; contains the point 1 - 2, 642
119. G1x2 = 1x + 32 2 1x - 22
(a) Identify the x-intercepts of the graph of G.
(b) What are the x-intercepts of the graph
y = G1x + 32 ?
120. h1x2 = 1x + 22 1x - 42 3
(a) Identify the x-intercepts of the graph of h.
(b) What are the x-intercepts of the graph
y = h1x - 22 ?
of
of
SECTION 5.1 Polynomial Functions and Models
341
Applications and Extensions
121. Hurricanes In 2012, Hurricane Sandy struck the East Coast
of the United States, killing 147 people and causing an
estimated $75 billion in damage. With a gale diameter of
about 1000 miles, it was the largest ever to form over the
Atlantic Basin. The accompanying data represent the number
of major hurricane strikes in the Atlantic Basin (category 3, 4,
or 5) each decade from 1921 to 2010.
Decade, x
Major Hurricanes Striking
Atlantic Basin, H
1921–1930, 1
17
1931–1940, 2
16
1941–1950, 3
29
1951–1960, 4
33
1961–1970, 5
27
1971–1980, 6
16
1981–1990, 7
16
1991–2000, 8
27
2001–2010, 9
33
Year, t
Percent below
Poverty Level, p
1990, 1
13.5
2001, 12
11.7
1991, 2
14.2
2002, 13
12.1
1992, 3
14.8
2003, 14
12.5
1993, 4
15.1
2004, 15
12.7
1994, 5
14.5
2005, 16
12.6
1995, 6
13.8
2006, 17
12.3
1996, 7
13.7
2007, 18
12.5
1997, 8
13.3
2008, 19
13.2
1998, 9
12.7
2009, 20
14.3
1999, 10
11.9
2010, 21
15.3
2000, 11
11.3
2011, 22
15.9
Source: U.S. Census Bureau
123. Temperature The following data represent the temperature
T (°Fahrenheit) in Kansas City, Missouri, x hours after
midnight on April 15, 2014.
Source: National Oceanic & Atmospheric
Administration
Temperature (°F), T
3
36.1
6
32.0
9
39.0
12
46.2
15
52.0
18
55.0
21
52.0
24
48.9
(a) Draw a scatter diagram of the data. Comment on the type
of relation that may exist between the two variables.
(b) Find the average rate of change in temperature from
9 am to 12 noon.
(c) What is the average rate of change in temperature from
3 pm to 6 pm ?
(d) Decide on a function of best fit to these data (linear,
quadratic, or cubic) and use this function to predict the
temperature at 5 pm.
(e) With a graphing utility, draw a scatter diagram of the
data and then graph the function of best fit on the
scatter diagram.
(f) Interpret the y-intercept.
124. Future Value of Money Suppose that you make deposits
of $500 at the beginning of every year into an Individual
Retirement Account (IRA) earning interest r (expressed as
a decimal). At the beginning of the first year, the value of
the account will be $500; at the beginning of the second year,
the value of the account, will be
+500 + +500r
+
Value of 1st deposit
+500 = +50011 + r2 + +500 = 500r + 1000
6
122. Poverty Rates The following data represent the percentage
of families in the United States whose income is below the
poverty level.
(a) With a graphing utility, draw a scatter diagram of the
data. Comment on the type of relation that appears to
exist between the two variables.
(b) Decide on a function of best fit to these data (linear,
quadratic, or cubic), and use this function to predict the
percentage of U.S. families that were below the poverty
level in 2012 (t = 23). Compare your prediction to the
actual value of 15.9.
(c) Draw the function of best fit on the scatter diagram
drawn in part (a).
Hours after Midnight, x
Source: The Weather Underground
e
(a) Draw a scatter diagram of the data. Comment on
the type of relation that may exist between the two
variables.
(b) Use a graphing utility to find the cubic function of best
fit that models the relation between decade and number
of major hurricanes.
(c) Use the model found in part (b) to predict the number
of major hurricanes that struck the Atlantic Basin
between 1961 and 1970.
(d) With a graphing utility, draw a scatter diagram of the
data and then graph the cubic function of best fit on the
scatter diagram.
(e) Concern has risen about the increase in the number
and intensity of hurricanes, but some scientists
believe this is just a natural fluctuation that could last
another decade or two. Use your model to predict
the number of major hurricanes that will strike the
Atlantic Basin between 2011 and 2020. Is your result
reasonable?
Percent below
Poverty Level, p
Year, t
Value of 2nd deposit
342
CHAPTER 5 Polynomial and Rational Functions
(a) Verify that the value of the account at the beginning of
the third year is T 1r2 = 500r 2 + 1500r + 1500.
(b) The account value at the beginning of the fourth year is
F1r2 = 500r 3 + 2000r 2 + 3000r + 2000. If the annual
rate of interest is 5% = 0.05, what will be the value of
the account at the beginning of the fourth year?
125. A Geometric Series In calculus, you will learn that certain
functions can be approximated by polynomial functions. We
will explore one such function now.
(a) Using a graphing utility, create a table of values with
1
Y1 = f 1x2 =
and Y2 = g2 1x2 = 1 + x + x2 + x3
1 - x
for - 1 6 x 6 1 with ∆Tbl = 0.1.
(b) Using a graphing utility, create a table of values with
1
and
Y1 = f 1x2 =
1-x
Y2 = g3 1x2 = 1 + x + x2 + x3 + x4
for - 1 6 x 6 1 with ∆Tbl = 0.1.
(c) Using a graphing utility, create a table of values with
1
Y1 = f 1x2 =
and
1 - x
Y2 = g4 1x2 = 1 + x + x2 + x3 + x4 + x5
for - 1 6 x 6 1 with ∆Tbl = 0.1.
(d) What do you notice about the values of the function as
more terms are added to the polynomial? Are there some
values of x for which the approximations are better?
Explaining Concepts: Discussion and Writing
126. Can the graph of a polynomial function have no y-intercept?
Can it have no x-intercepts? Explain.
132. The illustration shows the graph of a polynomial function.
127. Write a few paragraphs that provide a general strategy for
graphing a polynomial function. Be sure to mention the
following: degree, intercepts, end behavior, and turning points.
y
128. Make up a polynomial that has the following characteristics:
crosses the x-axis at - 1 and 4, touches the x-axis at 0 and 2,
and is above the x-axis between 0 and 2. Give your polynomial
to a fellow classmate and ask for a written critique.
129. Make up two polynomials, not of the same degree, with the
following characteristics: crosses the x-axis at - 2, touches
the x-axis at 1, and is above the x-axis between - 2 and 1.
Give your polynomials to a fellow classmate and ask for a
written critique.
130. The graph of a polynomial function is always smooth and
continuous. Name a function studied earlier that is smooth
but not continuous. Name one that is continuous but not
smooth.
131. Which of the following statements are true regarding the
graph of the cubic polynomial f 1x2 = x3 + bx2 + cx + d?
(Give reasons for your conclusions.)
(a) It intersects the y-axis in one and only one point.
(b) It intersects the x-axis in at most three points.
(c) It intersects the x-axis at least once.
(d) For x very large, it behaves like the graph of y = x3.
(e) It is symmetric with respect to the origin.
(f) It passes through the origin.
x
(a)
(b)
(c)
(d)
(e)
(f)
Is the degree of the polynomial even or odd?
Is the leading coefficient positive or negative?
Is the function even, odd, or neither?
Why is x2 necessarily a factor of the polynomial?
What is the minimum degree of the polynomial?
Formulate five different polynomials whose graphs
could look like the one shown. Compare yours to those
of other students. What similarities do you see? What
differences?
133. Design a polynomial function with the following
characteristics: degree 6; four distinct real zeros, one of
multiplicity 3; y-intercept 3; behaves like y = - 5x6 for large
values of x . Is this polynomial unique? Compare your
polynomial with those of other students. What terms will be
the same as everyone else’s? Add some more characteristics,
such as symmetry or naming the real zeros. How does this
modify the polynomial?
Retain Your Knowledge
Problems 134–137 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in
your mind so that you are better prepared for the final exam.
134. Find the equation of the line that contains the point 12, - 32 and is perpendicular to the line 5x - 2y = 6.
x - 3
.
135. Find the domain of the function h1x2 =
x + 5
136. Use the quadratic formula to find the zeros of the function f 1x2 = 4x2 + 8x - 3.
137. Solve: 5x - 3 = 7.
‘Are You Prepared?’ Answers
1. 1 - 2, 02, 12, 02, 10, 92
5. True
6. (5, 0); 5
2. Yes; 3 3. Down; 4
4. Local maximum value 6.48 at x = - 0.67; local minimum value - 3 at x = 2
SECTION 5.2 Properties of Rational Functions
343
5.2 Properties of Rational Functions
PREPARING FOR THIS SECTION Before getting started, review the following:
r Graph of f 1x2 =
r Rational Expressions (Chapter R, Section R.7,
pp. 62–69)
r Polynomial Division (Chapter R, Section R.4,
pp. 44–47)
1
(Section 2.2, Example 12, p. 164)
x
r Graphing Techniques: Transformations (Section 3.5,
pp. 247–256)
Now Work the ‘Are You Prepared?’ problems on page 350.
OBJECTIVES 1 Find the Domain of a Rational Function (p. 343)
2 Find the Vertical Asymptotes of a Rational Function (p. 346)
3 Find the Horizontal or Oblique Asymptote of a Rational Function (p. 348)
Ratios of integers are called rational numbers. Similarly, ratios of polynomial
functions are called rational functions. Examples of rational functions are
R 1x2 =
DEFINITION
x2 - 4
x + x + 1
2
F1x2 =
x3
x - 4
2
G1x2 =
3x2
x - 1
4
A rational function is a function of the form
R 1x2 =
p 1x2
q 1x2
where p and q are polynomial functions and q is not the zero polynomial. The
domain of a rational function is the set of all real numbers except those for
which the denominator q is 0.
1 Find the Domain of a Rational Function
EXAMPL E 1
Finding the Domain of a Rational Function
2x2 - 4
is the set of all real numbers x except - 5; that
x + 5
is, the domain is 5 x x ≠ - 56 .
(a) The domain of R 1x2 =
(b) The domain of R 1x2 =
1
1
=
is the set of all real numbers x
1x
+
22
1x - 22
x - 4
except - 2 and 2; that is, the domain is 5 x x ≠ - 2, x ≠ 26 .
(c) The domain of R 1x2 =
2
x3
is the set of all real numbers.
x2 + 1
x2 - 1
is the set of all real numbers x except 1; that is,
x - 1
the domain is 5 x x ≠ 16 .
(d) The domain of R 1x2 =
r
x2 - 1
Although
reduces to x + 1, it is important to observe that the functions
x - 1
R 1x2 =
x2 - 1
x - 1
and f1x2 = x + 1
are not equal, since the domain of R is 5 x x ≠ 16 and the domain of f is the set
of all real numbers.
Now Work
PROBLEM
17
344
CHAPTER 5 Polynomial and Rational Functions
WARNING The domain of a rational
function must be found before writing the
function in lowest terms.
■
EX AMPLE 2
p 1x2
is a rational function, and if p and q have no common factors,
q 1x2
then the rational function R is said to be in lowest terms. For a rational function
p 1x2
R 1x2 =
in lowest terms, the real zeros, if any, of the numerator in the domain
q 1x2
of R are the x-intercepts of the graph of R and so will play a major role in the graph
of R. The real zeros of the denominator of R [that is, the numbers x, if any, for which
q 1x2 = 0], although not in the domain of R, also play a major role in the graph of R.
1
We have already discussed the properties of the rational function y = .
x
(Refer to Example 12, page 164). The next rational function that we take up is
1
H1x2 = 2 .
x
If R 1x2 =
Graphing y =
1
x2
1
.
x2
Analyze the graph of H1x2 =
Solution
1
is the set of all real numbers x except 0. The graph has
x2
no y-intercept, because x can never equal 0. The graph has no x-intercept because
the equation H1x2 = 0 has no solution. Therefore, the graph of H will not cross or
touch either of the coordinate axes. Because
The domain of H1x2 =
H1 - x2 =
Table 7
x
H(x) =
1
x2
1
2
4
1
100
10,000
1
10,000
100,000,000
1
1
2
100
10,000
H is an even function, so its graph is symmetric with respect to the y-axis.
1
Table 7 shows the behavior of H1x2 = 2 for selected positive numbers x. (We
x
use symmetry to obtain the graph of H when x 6 0.) From the first three rows
of Table 7, we see that as the values of x approach (get closer to) 0, the values of
H1x2 become larger and larger positive numbers, so H is unbounded in the positive
direction. In calculus we use limit notation, lim H1x2 = q , which is read “the limit
xS0
of H1x2 as x approaches zero equals infinity,” to mean that H1x2 S q as x S 0.
Look at the last four rows of Table 7. As x S q , the values of H1x2
approach 0 (the end behavior of the graph). In calculus, this is expressed by writing
lim H1x2 = 0. Figure 25 shows the graph. Notice the use of red dashed lines to
x Sq
convey the ideas discussed above.
x0
y
5
( 1–2 , 4)
1
4
1
10,000
1
100,000,000
(1, 1)
1–
4
Figure 25 H(x) =
( 1–2 , 4)
(1, 1)
(2, 1–4 )
(2, )
3 x y0
y 0 3
EX AMPLE 3
1
1
= 2 = H1x2
2
1 - x2
x
1
x2
r
Using Transformations to Graph a Rational Function
Graph the rational function: R 1x2 =
1
+ 1
1x - 22 2
345
SECTION 5.2 Properties of Rational Functions
The domain of R is the set of all real numbers except x = 2. To graph R, start with
1
the graph of y = 2 . See Figure 26 for the steps.
x
Solution
x⫽2
x⫽0
x⫽2
y
y
y
3
3
(3, 2)
(⫺1, 1)
1
(1, 1)
y ⫽ 0 ⫺2
3
y⫽0
x
x
5
Replace x by x ⫺ 2;
shift right
2 units
(a) y ⫽
Figure 26
(1, 2)
y⫽1
(3, 1)
(1, 1)
1
x2
PROBLEMS
x
Add 1;
shift up
1 unit
(b) y ⫽
Now Work
5
1
(x – 2)2
35(a)
(c) y ⫽
1
⫹1
(x ⫺ 2)2
35(b)
AND
r
Asymptotes
Let’s investigate the roles of the vertical line x = 2 and the horizontal line y = 1
in Figure 26(c).
1
First, we look at the end behavior of R 1x2 =
+ 1. Table 8(a) shows
1x - 22 2
the values of R at x = 10, 100, 1000, and 10,000. Note that as x becomes unbounded
in the positive direction, the values of R approach 1, so lim R 1x2 = 1. From Table 8(b)
x Sq
we see that as x becomes unbounded in the negative direction, the values of R
also approach 1, so lim R 1x2 = 1.
xS -q
Even though x = 2 is not in the domain of R, the behavior of the graph of R
near x = 2 is important. Table 8(c) shows the values of R at x = 1.5, 1.9, 1.99, 1.999,
and 1.9999. We see that as x approaches 2 for x 6 2, denoted x S 2 - , the values of
R are increasing without bound, so lim- R 1x2 = q . From Table 8(d), we see that
xS2
as x approaches 2 for x 7 2, denoted x S 2 + , the values of R are also increasing
without bound, so lim+ R 1x2 = q .
Table 8
xS2
x
R(x)
x
R(x)
x
R(x)
x
R(x)
10
1.0156
- 10
1.0069
1.5
5
2.5
5
100
1.0001
- 100
1.0001
1.9
101
2.1
101
1000
1.000001
- 1000
1.000001
1.99
10,001
2.01
10,001
10,000
1.00000001
- 10,000
1.00000001
1.999
1,000,001
2.001
1,000,001
1.9999
100,000,001
2.0001
100,000,001
(a)
(b)
(c)
(d)
The vertical line x = 2 and the horizontal line y = 1 are called asymptotes of
the graph of R.
DEFINITION
Let R denote a function.
If, as x S - q or as x S q , the values of R 1x2 approach some fixed
number L, then the line y = L is a horizontal asymptote of the graph of R.
[Refer to Figures 27(a) and (b) on page 346.]
If, as x approaches some number c, the values R 1x2 S q [that is,
R 1x2 S - q or R 1x2 S q ], then the line x = c is a vertical asymptote of the
graph of R. [Refer to Figures 27(c) and (d).]
346
CHAPTER 5 Polynomial and Rational Functions
x5c
y
y
y
y
x5c
y 5 R(x )
y5L
y5L
x
D
y 5 R(x )
End behavior:
As x → `, the values of
R(x ) approach L [ xlim
R(x) 5 L].
→`
That is, the points on the graph
of R are getting closer to
the line y 5 L; y 5 L is a
horizontal asymptote.
E
End behavior:
As x → 2`, the values
of R(x) approach L
lim R (x) 5 L]. That is, the
[ x→
2`
points on the graph of R
are getting closer to the line
y 5 L; y 5 L is a horizontal
asymptote.
x
x
x
F
As x approaches c, the
values of ⏐R (x)⏐→ `
[ xlim
→ c2 R(x) 5 `;
lim
x → c 1R(x) 5 `]. That is,
the points on the graph
of R are getting closer to
the line x 5 c ; x 5 c is a
vertical asymptote.
G
As x approaches c, the
values of ⏐R(x)⏐→ `
[ xlim
→ c2 R(x) 5 2`;
lim R(x) 5 `]. That is,
x → c1
the points on the graph
of R are getting closer to
the line x 5 c; x 5 c is a
vertical asymptote.
Figure 27
y
x
Figure 28 Oblique asymptote
A horizontal asymptote, when it occurs, describes the end behavior of the
graph as x S q or as x S - q . The graph of a function may intersect a horizontal
asymptote.
A vertical asymptote, when it occurs, describes the behavior of the graph when
x is close to some number c. The graph of a rational function will never intersect a
vertical asymptote.
There is a third possibility. If, as x S - q or as x S q , the value of a rational
function R 1x2 approaches a linear expression ax + b, a ≠ 0, then the line
y = ax + b, a ≠ 0, is an oblique (or slant) asymptote of R. Figure 28 shows an oblique
asymptote. An oblique asymptote, when it occurs, describes the end behavior of the
graph. The graph of a function may intersect an oblique asymptote.
Now Work
PROBLEMS
27
AND
35(c)
2 Find the Vertical Asymptotes of a Rational Function
p 1x2
, in lowest terms, are
q 1x2
located at the real zeros of the denominator q 1x2. Suppose that r is a real zero of q,
so x - r is a factor of q. As x approaches r, symbolized as x S r, the values of x - r
approach 0, causing the ratio to become unbounded; that is, R 1x2 S q . Based on
the definition, we conclude that the line x = r is a vertical asymptote.
The vertical asymptotes of a rational function R 1x2 =
THEOREM
WARNING If a rational function is not
in lowest terms, an application of this
theorem may result in an incorrect listing
of vertical asymptotes.
■
EX AMPLE 4
Locating Vertical Asymptotes
p 1x2
, in lowest terms, will have a vertical asymptote
A rational function R 1x2 =
q 1x2
x = r if r is a real zero of the denominator q. That is, if x - r is a factor of the
p 1x2
denominator q of a rational function R 1x2 =
, in lowest terms, R will
q 1x2
have the vertical asymptote x = r.
Finding Vertical Asymptotes
Find the vertical asymptotes, if any, of the graph of each rational function.
x + 3
x - 1
x2
(c) H1x2 = 2
x + 1
(a) F1x2 =
(b) R 1x2 =
x
x - 4
x2 - 9
(d) G1x2 = 2
x + 4x - 21
2
SECTION 5.2 Properties of Rational Functions
Solution
WARNING In Example 4(a), the vertical
asymptote is x = 1. Do not say that
the vertical asymptote is 1.
■
347
(a) F is in lowest terms, and the only zero of the denominator is 1. The line x = 1 is
the vertical asymptote of the graph of F.
(b) R is in lowest terms, and the zeros of the denominator x2 - 4 are - 2 and 2. The
lines x = - 2 and x = 2 are the vertical asymptotes of the graph of R.
(c) H is in lowest terms, and the denominator has no real zeros because the equation
x2 + 1 = 0 has no real solutions. The graph of H has no vertical asymptotes.
(d) Factor the numerator and denominator of G1x2 to determine whether it is in
lowest terms.
1x + 32 1x - 32
x2 - 9
x + 3
=
=
x ≠ 3
1x + 72 1x - 32
x + 7
x + 4x - 21
The only zero of the denominator of G1x2 in lowest terms is - 7. The line
x = - 7 is the only vertical asymptote of the graph of G.
G1x2 =
2
r
As Example 4 points out, rational functions can have no vertical asymptotes,
one vertical asymptote, or more than one vertical asymptote.
Multiplicity and Vertical Asymptotes
Recall from Figure 15 in Section 5.1 that the end behavior of a polynomial function
is always one of four types. For polynomials of odd degree, the ends of the graph
go in opposite directions (one up and one down), whereas for polynomials of even
degree, the ends go in the same direction (both up or both down).
For a rational function in lowest terms, the multiplicities of the zeros in the
denominator can be used in a similar fashion to determine the behavior of the graph
around each vertical asymptote. Consider the following four functions, each with a
single vertical asymptote, x = 2.
1
1
1
1
R2 1x2 = R3 1x2 =
R4 1x2 = R1 1x2 =
x - 2
x - 2
1x - 22 2
1x - 22 2
Figure 29 shows the graphs of each function. The graphs of R1 and R2 are
1
transformations of the graph of y = , and the graphs of R3 and R4 are
x
1
transformations of the graph of y = 2 .
x
Based on Figure 29, we can make the following conclusions:
r If the multiplicity of the zero that gives rise to a vertical asymptote is odd, the graph
approaches q on one side of the vertical asymptote and approaches - q on
the other side.
r If the multiplicity of the zero that gives rise to the vertical asymptote is even,
the graph approaches either q or - q on both sides of the vertical asymptote.
These results are true in general and will be helpful when graphing rational functions
in the next section.
y
6
y
6
R1(x)
−1
5
−6
x2
(a) Odd multiplicity
lim- R1(x) = - q
xS2
lim+ R1(x) = q
xS2
Figure 29
y
6
x2
y
6
R3(x)
x2
y0
y0
y0
y0
x
x
x
x
−1
5
−6
R2(x)
(b) Odd multiplicity
lim- R2(x) = q
xS2
lim+ R2(x) = - q
xS2
−1
5
−6
x2
(c) Even multiplicity
lim- R3(x) = q
xS2
lim+ R3(x) = q
xS2
−1
5
−6
R4(x)
(d) Even multiplicity
lim- R4(x) = - q
xS2
lim R4(x) = - q
xS2 +
348
CHAPTER 5 Polynomial and Rational Functions
3 Find the Horizontal or Oblique Asymptote
of a Rational Function
To find horizontal or oblique asymptotes, we need to know how the value of the
function behaves as x S - q or as x S q . That is, we need to determine the end
behavior of the function. This can be done by examining the degrees of the numerator
and denominator, and the respective power functions that each resembles. For
example, consider the rational function
R 1x2 =
3x - 2
5x - 7x + 1
2
The degree of the numerator, 1, is less than the degree of the denominator, 2.
When x is very large, the numerator of R can be approximated by the power
function y = 3x, and the denominator can be approximated by the power function
y = 5x2. This means
R 1x2 =
3x - 2
3x
3
S0
≈
=
2
5x c
5x - 7x + 1 c 5x
2
For |x| very large
As x S - q or x S q
which shows that the line y = 0 is a horizontal asymptote. This result is true for all
rational functions that are proper (that is, the degree of the numerator is less than
the degree of the denominator). If a rational function is improper (that is, if the
degree of the numerator is greater than or equal to the degree of the denominator),
there could be a horizontal asymptote, an oblique asymptote, or neither. The following
summary details how to find horizontal or oblique asymptotes.
Finding a Horizontal or Oblique Asymptote of a Rational Function
Consider the rational function
R 1x2 =
p 1x2
anxn + an - 1xn - 1 + g + a1x + a0
=
q 1x2
bmxm + bm - 1xm - 1 + g + b1x + b0
in which the degree of the numerator is n and the degree of the denominator
is m.
1. If n 6 m (the degree of the numerator is less than the degree of the
denominator), the line y = 0 is a horizontal asymptote.
2. If n = m (the degree of the numerator equals the degree of the
an
denominator), the line y =
is a horizontal asymptote. (That is, the
bm
horizontal asymptote equals the ratio of the leading coefficients.)
3. If n = m + 1 (the degree of the numerator is one more than the degree of
the denominator), the line y = ax + b is an oblique asymptote, which is
the quotient found using long division.
4. If n Ú m + 2 (the degree of the numerator is two or more greater than the
degree of the denominator), there are no horizontal or oblique asymptotes.
The end behavior of the graph will resemble the power function
an n - m
y =
x
.
bm
Note: A rational function will never have both a horizontal asymptote and an
oblique asymptote. A rational function may have neither a horizontal nor an
oblique asymptote.
We illustrate each of the possibilities in Examples 5 through 8.
SECTION 5.2 Properties of Rational Functions
EXAMPL E 5
349
Finding a Horizontal Asymptote
Find the horizontal asymptote, if one exists, of the graph of
R 1x2 =
Solution
4x3 - 5x + 2
7x5 + 2x4 - 3x
Since the degree of the numerator, 3, is less than the degree of the denominator, 5,
the rational function R is proper. The line y = 0 is a horizontal asymptote of the
graph of R.
r
EXAMPL E 6
Finding a Horizontal or Oblique Asymptote
Find the horizontal or oblique asymptote, if one exists, of the graph of
H1x2 =
Solution
3x4 - x2
x3 - x2 + 1
Since the degree of the numerator, 4, is exactly one greater than the degree of the
denominator, 3, the rational function H has an oblique asymptote. Find the
asymptote by using long division.
3x + 3
- x2
x3 - x2 + 1 ) 3x4
3x4 - 3x3
+ 3x
3x3 - x2 - 3x
3x3 - 3x2
+ 3
2x2 - 3x - 3
As a result,
H1x2 =
3x4 - x2
2x2 - 3x - 3
=
3x
+
3
+
x3 - x2 + 1
x3 - x2 + 1
As x S - q or as x S q ,
2x2 - 3x - 3
2x2
2
≈
= S0
3
2
3
x
x - x + 1
x
As x S - q or as x S q , we have H1x2 S 3x + 3. The graph of the rational
function H has an oblique asymptote y = 3x + 3. Put another way, as x S { q , the
graph of H will behave like the graph of y = 3x + 3.
r
EXAMPL E 7
Finding a Horizontal or Oblique Asymptote
Find the horizontal or oblique asymptote, if one exists, of the graph of
R 1x2 =
Solution
8x2 - x + 2
4x2 - 1
Since the degree of the numerator, 2, equals the degree of the denominator, 2,
the rational function R has a horizontal asymptote equal to the ratio of the leading
coefficients.
y =
an
8
= = 2
bm
4
To see why the horizontal asymptote equals the ratio of the leading coefficients,
investigate the behavior of R as x S - q or as x S q . When x is very large,
the numerator of R can be approximated by the power function y = 8x2, and the
350
CHAPTER 5 Polynomial and Rational Functions
denominator can be approximated by the power function y = 4x2. This means that
as x S - q or as x S q ,
R 1x2 =
8
8x2
8x2 - x + 2
= = 2
≈
2
2
4
4x - 1
4x
The graph of the rational function R has a horizontal asymptote y = 2. The
graph of R will behave like y = 2 as x S { q .
r
EX AMPLE 8
Finding a Horizontal or Oblique Asymptote
Find the horizontal or oblique asymptote, if one exists, of the graph of
G1x2 =
Solution
2x5 - x3 + 2
x3 - 1
Since the degree of the numerator, 5, is greater than the degree of the denominator, 3,
by more than one, the rational function G has no horizontal or oblique asymptote.
The end behavior of the graph will resemble the power function y = 2x5-3 = 2x2.
To see why this is the case, investigate the behavior of G as x S - q or as
x S q . When x is very large, the numerator of G can be approximated by the
power function y = 2x5, and the denominator can be approximated by the power
function y = x3. This means as x S - q or as x S q ,
G1x2 =
2x5
2x5 - x3 + 2
≈
= 2x5 - 3 = 2x2
x3 - 1
x3
Since this is not linear, the graph of G has no horizontal or oblique asymptote.
The graph of G will behave like y = 2x2 as x S { q .
r
Now Work
PROBLEMS
45, 47,
AND
49
5.2 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. True or False The quotient of two polynomial expressions
is a rational expression. (pp. 62–69)
3. Graph y =
2. What are the quotient and remainder when 3x4 - x2 is
divided by x3 - x2 + 1. (pp. 44–47)
4. Graph y = 21x + 12 2 - 3 using transformations.
1
. (p. 164)
x
(pp. 247–256)
Concepts and Vocabulary
5. True or False The domain of every rational function is the
set of all real numbers.
6. If, as x S - q or as x S q , the values of R 1x2 approach
some fixed number L, then the line y = L is a
of the graph of R.
7. If, as x approaches some number c, the values of
R 1x2 S q , then the line x = c is a
of the graph of R.
8. For a rational function R, if the degree of the numerator is
less than the degree of the denominator, then R is
.
9. True or False The graph of a rational function may intersect
a horizontal asymptote.
10. True or False The graph of a rational function may intersect
a vertical asymptote.
11. If a rational function is proper, then
asymptote.
is a horizontal
12. True or False If the degree of the numerator of a rational
function equals the degree of the denominator, then the
ratio of the leading coefficients gives rise to the horizontal
asymptote.
p1x2
13. If R 1x2 =
is a rational function and if p and q have no
q1x2
common factors, then R is
.
(a) improper
(c) undefined
(b) proper
(d) in lowest terms
14. Which type of asymptote, when it occurs, describes the
behavior of a graph when x is close to some number?
(a) vertical (b) horizontal (c) oblique (d) all of these
351
SECTION 5.2 Properties of Rational Functions
Skill Building
In Problems 15–26, find the domain of each rational function.
15. R 1x2 =
4x
x - 3
16. R 1x2 =
18. G1x2 =
6
1x + 32 14 - x2
19. F 1x2 =
21. R 1x2 =
x
x - 8
22. R 1x2 =
x
x - 1
23. H1x2 =
24. G1x2 =
x - 3
x4 + 1
25. R 1x2 =
31x2 - x - 62
26. F 1x2 =
3
In Problems 27–32, use the graph shown to find
(a) The domain and range of each function
(d) Vertical asymptotes, if any
27.
5x2
3 + x
3x1x - 12
20. Q1x2 =
2
2x - 5x - 3
4
2
41x - 92
(b) The intercepts, if any
(e) Oblique asymptotes, if any
28.
y
17. H1x2 =
- x11 - x2
3x2 + 5x - 2
3x2 + x
x2 + 4
- 21x2 - 42
31x2 + 4x + 42
(c) Horizontal asymptotes, if any
29.
y
4
- 4x2
1x - 22 1x + 42
y
3
3
(0, 2)
4 x
–4
3
(1, 2)
(1, 0)
(1, 0)
3 x
3
–4
3 x
3
3
30.
31.
32.
y
y
y
3
(1, 2) 3
3
(1, 1)
3
3
3 x
3
3
3 x
3
(1, 2)
3
3
In Problems 33–44, (a) graph the rational function using transformations, (b) use the final graph to find the domain and range, and
(c) use the final graph to list any vertical, horizontal, or oblique asymptotes.
33. F 1x2 = 2 +
37. H1x2 =
1
x
-2
x + 1
41. G1x2 = 1 +
2
1x - 32 2
34. Q1x2 = 3 +
38. G1x2 =
1
x2
2
1x + 22 2
42. F 1x2 = 2 -
1
x + 1
35. R 1x2 =
1
1x - 12 2
36. R 1x2 =
3
x
39. R 1x2 =
-1
x + 4x + 4
40. R 1x2 =
1
+ 1
x - 1
43. R 1x2 =
x2 - 4
x2
44. R 1x2 =
x - 4
x
2
x
352
CHAPTER 5 Polynomial and Rational Functions
In Problems 45–56, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.
45. R 1x2 =
3x
x + 4
46. R 1x2 =
3x + 5
x - 6
47. H1x2 =
x3 - 8
x - 5x + 6
48. G1x2 =
x3 + 1
x - 5x - 14
49. T 1x2 =
x3
x - 1
50. P 1x2 =
4x2
x - 1
51. Q1x2 =
2x2 - 5x - 12
3x2 - 11x - 4
52. F 1x2 =
x2 + 6x + 5
2x2 + 7x + 5
53. R 1x2 =
6x2 + 7x - 5
3x + 5
54. R 1x2 =
8x2 + 26x - 7
4x - 1
55. G1x2 =
x4 - 1
x2 - x
56. F 1x2 =
x4 - 16
x2 - 2x
4
3
2
2
Applications and Extensions
57. Gravity In physics, it is established that the acceleration due
to gravity, g (in meters/sec2), at a height h meters above sea
level is given by
60. Newton’s Method In calculus you will learn that if
p1x2 = anxn + an-1xn-1 + g + a1x + a0
is a polynomial function, then the derivative of p1x2 is
g1h2 =
3.99 * 1014
16.374 * 10 + h2
6
2
where 6.374 * 106 is the radius of Earth in meters.
(a) What is the acceleration due to gravity at sea level?
(b) The Willis Tower in Chicago, Illinois, is 443 meters tall.
What is the acceleration due to gravity at the top of the
Willis Tower?
(c) The peak of Mount Everest is 8848 meters above sea
level. What is the acceleration due to gravity on the peak
of Mount Everest?
(d) Find the horizontal asymptote of g1h2 .
(e) Solve g1h2 = 0. How do you interpret your answer?
58. Population Model A rare species of insect was discovered
in the Amazon Rain Forest. To protect the species,
environmentalists declared the insect endangered and
transplanted the insect into a protected area.The population P
of the insect t months after being transplanted is
P 1t2 =
5011 + 0.5t2
2 + 0.01t
(a) How many insects were discovered? In other words,
what was the population when t = 0?
(b) What will the population be after 5 years?
(c) Determine the horizontal asymptote of P 1t2 . What
is the largest population that the protected area can
sustain?
59. Resistance in Parallel Circuits From Ohm’s Law for circuits,
it follows that the total resistance Rtot of two components
hooked in parallel is given by the equation
Rtot =
R1R2
R1 + R2
where R1 and R2 are the individual resistances.
(a) Let R1 = 10 ohms, and graph Rtot as a function of R2.
(b) Find and interpret any asymptotes of the graph obtained
in part (a).
(c) If R2 = 21R1, what value of R1 will yield an Rtot of
17 ohms?
p′ 1x2 = nanxn-1 + 1n - 12an-1xn-2 + g + 2a2x + a1
Newton’s Method is an efficient method for approximating
the x-intercepts (or real zeros) of a function, such as p1x2 .
The following steps outline Newton’s Method.
STEP 1: Select an initial value x0 that is somewhat close to
the x-intercept being sought.
STEP 2: Find values for x using the relation
xn + 1 = xn -
p1xn 2
p′ 1xn 2
n = 1, 2, c
until you get two consecutive values xn and xn + 1
that agree to whatever decimal place accuracy you
desire.
STEP 3: The approximate zero will be xn + 1.
Consider the polynomial p1x2 = x3 - 7x - 40.
(a) Evaluate p152 and p1 - 32 .
(b) What might we conclude about a zero of p? Explain.
(c) Use Newton’s Method to approximate an x-intercept, r,
- 3 6 r 6 5, of p1x2 to four decimal places.
(d) Use a graphing utility to graph p1x2 and verify your
answer in part (c).
(e) Using a graphing utility, evaluate p1r2 to verify your
result.
61. Exploration The standard form of the rational
mx + b
function R 1x2 =
, where c ≠ 0,
cx + d
1
is R 1x2 = a¢
≤ + k. To write a rational function
x - h
in standard form requires long division.
2x + 3
(a) Write the rational function R 1x2 =
in
x - 1
standard form by writing R in the form
Quotient +
remainder
divisor
(b) Graph R using transformations.
(c) Determine the vertical asymptote and the horizontal
asymptote of R.
62. Exploration Repeat
Problem 61
- 6x + 16
function R 1x2 =
.
2x - 7
for
the
rational
SECTION 5.3 The Graph of a Rational Function
353
Explaining Concepts: Discussion and Writing
63. If the graph of a rational function R has the vertical
asymptote x = 4, the factor x - 4 must be present in the
denominator of R. Explain why.
64. If the graph of a rational function R has the horizontal
asymptote y = 2, the degree of the numerator of R equals
the degree of the denominator of R. Explain why.
65. The graph of a rational function cannot have both a horizontal
and an oblique asymptote? Explain why.
66. Make up a rational function that has y = 2x + 1 as an
oblique asymptote. Explain the methodology that you used.
Retain Your Knowledge
Problems 67–70 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
x
2
68. Solve: 13x - 72 + 1 = - 2
67. Find the equation of a vertical line passing through the
5
4
point 15, - 32 .
69. Determine whether the graph of the equation
2x3 - xy2 = 4 is symmetric with respect to the x-axis,
the y-axis, the origin, or none of these.
70. What are the points of intersection of the graphs of the
functions f 1x2 = - 3x + 2 and g1x2 = x2 - 2x - 4?
‘Are You Prepared?’ Answers
1. True
2. Quotient: 3x + 3; remainder: 2x2 - 3x - 3
y
3.
y
3
4.
2
(1, 1)
2
3
2 x
3 x
(0,1)
(1, 1)
2
(1,3)
3
5.3 The Graph of a Rational Function
PREPARING FOR THIS SECTION Before getting started, review the following:
r Intercepts Section 2.2, pp. 159–160
Now Work the ‘Are You Prepared?’ problem on page 365.
OBJECTIVES 1 Analyze the Graph of a Rational Function (p. 353)
2 Solve Applied Problems Involving Rational Functions (p. 364)
1 Analyze the Graph of a Rational Function
We commented earlier that calculus provides the tools required to graph a polynomial
function accurately. The same holds true for rational functions. However, we can
gather together quite a bit of information about their graphs to get an idea of the
general shape and position of the graph.
EXAMPL E 1
How to Analyze the Graph of a Rational Function
Analyze the graph of the rational function: R 1x2 =
Step-by-Step Solution
Step 1: Factor the numerator and
denominator of R. Find the domain
of the rational function.
R 1x2 =
x - 1
x2 - 4
x - 1
x - 1
=
2
1x + 22 1x - 22
x - 4
The domain of R is {x x ≠ - 2, x ≠ 2}.
354
CHAPTER 5 Polynomial and Rational Functions
Step 2: Write R in lowest terms.
Step 3: Find and plot the
intercepts of the graph. Use
multiplicity to determine the
behavior of the graph of R at each
x-intercept.
Because there are no common factors between the numerator and denominator,
R is in lowest terms.
1
1
. Plot the point ¢ 0, ≤. The
4
4
x-intercepts are found by determining the real zeros of the numerator of R written
in lowest terms. By solving x - 1 = 0, we find that the only real zero of the
numerator is 1, so the only x-intercept of the graph of R is 1. Plot the point (1, 0). The
multiplicity of 1 is odd, so the graph will cross the x-axis at x = 1.
Since 0 is in the domain of R, the y-intercept is R(0) =
Step 4: Find the vertical
asymptotes. Graph each vertical
asymptote using a dashed line.
Determine the behavior of the graph
on either side of each vertical
asymptote.
The vertical asymptotes are the zeros of the denominator with the rational
function in lowest terms. With R written in lowest terms, we find that the graph of
R has two vertical asymptotes: the lines x = - 2 and x = 2. See Figure 30(a). The
multiplicities of the zeros that give rise to the vertical asymptotes are both odd.
Therefore, the graph will approach q on one side of each vertical asymptote, and
will approach - q on the other side.
Step 5: Find the horizontal or
oblique asymptote, if one exists.
Find points, if any, at which
the graph of R intersects this
asymptote. Graph the asymptote
using a dashed line. Plot any points
at which the graph of R intersects
the asymptote.
Because the degree of the numerator is less than the degree of the denominator,
R is proper and the line y = 0 (the x-axis) is a horizontal asymptote of the graph.
To determine whether the graph of R intersects the horizontal asymptote, solve the
equation R 1x2 = 0:
x - 1
= 0
x2 - 4
x - 1 = 0
x = 1
The only solution is x = 1, so the graph of R intersects the horizontal asymptote at 11, 02.
Step 6: Use the zeros of the
numerator and denominator of R
to divide the x-axis into intervals.
Determine where the graph of R
is above or below the x-axis by
choosing a number in each interval
and evaluating R there. Plot the
points found.
The zero of the numerator, 1, and the zeros of the denominator, - 2 and 2, divide the
x-axis into four intervals:
1 - q , - 22
1 - 2, 12
11, 22
12, q 2
Now construct Table 9.
Table 9
–2
1
2
x
Interval
( - q , - 2)
( - 2, 1)
(1, 2)
(2, q )
Number chosen
-3
-1
3
2
3
Value of R
R( - 3) = - 0.8
R( - 1) =
2
3
Ra b = 2
7
R(3) = 0.4
Location of graph
Below x-axis
Above x-axis
Below x-axis
Above x-axis
Point on graph
( - 3, - 0.8)
2
a - 1, b
3
3 2
a ,- b
2 7
(3, 0.4)
2
3
Figure 30(a) shows the asymptotes, the points from Table 9, the y-intercept, and
the x-intercept.
Step 7: Use the results obtained in
Steps 1 through 6 to graph R.
r The graph crosses the x-axis at x = 1, changing from being above the x-axis
for x 6 1 to below it for x 7 1. Indicate this on the graph. See Figure 30(b).
r Since y = 0 (the x-axis) is a horizontal asymptote and the graph lies below the
x-axis for x 6 - 2, we can sketch a portion of the graph by placing a small arrow
to the far left and under the x-axis.
SECTION 5.3 The Graph of a Rational Function
355
r Since the line x = - 2 is a vertical asymptote and the graph lies below the x-axis
for x 6 - 2, we place an arrow well below the x-axis and approaching the line
x = - 2 from the left 1 lim -R 1x2 = - q 2 .
x S -2
r Since the graph approaches - q on one side of x = - 2, and - 2 is a zero of
odd multiplicity, the graph will approach q on the other side of x = - 2. That is,
lim +R 1x2 = q . Similar analysis leads to lim-R(x) = - q and lim+R(x) = q .
x S -2
Finally, lim R(x) = 0 and lim R(x) = 0.
xS -q
xS 2
xS 2
xS q
Figure 30(b) illustrates these conclusions and Figure 30(c) shows the graph of R.
x = −2
x=2
y
x = −2
3
(1, 0)
(−3, −0.8)
x = −2
(0, 1–4) ( 3–2 ,−2–7)
3
(1, 0)
xy = 0
−3
(−3, −0.8)
(0, 1–4) ( 3–2 ,−2–7)
(3, 0.4)
xy = 0
3
(−1, 2–3)
−3
x=2
(−3, −0.8)
(1, 0)
(0, 1–4) ( 3– ,−2–)
7
2
(3, 0.4)
3
xy=0
−3
−3
(a)
y
3
(−1, 2–3)
(3, 0.4)
−3
Figure 30
x=2
3
(−1, 2–3)
−3
y
(b)
r
(c)
Exploration
Graph the rational function: R(x) =
x - 1
x2 - 4
Result The analysis just completed in Example 1 helps us to set the viewing rectangle to obtain a
x - 1
complete graph. Figure 31(a) shows the graph of R(x) = 2
in connected mode, and Figure 31(b)
x - 4
shows it in dot mode. Notice in Figure 31(a) that the graph has vertical lines at x = - 2 and x = 2.
This is due to the fact that when a graphing utility is in connected mode, some will connect the
dots between consecutive pixels, and vertical lines may occur. We know that the graph of R does not
cross the lines x = - 2 and x = 2, since R is not defined at x = - 2 or x = 2. So, when graphing
rational functions, use dot mode if extraneous vertical lines are present in connected mode. Newer
graphing utilities may not have extraneous vertical lines in connected mode. See Figure 31(c).
4
4
24
4
24
4
4
24
4
24
Figure 31
24
Connected mode with extraneous
vertical lines
(a)
24
Dot mode
(b)
Connected mode without extraneous
vertical lines
(c)
356
CHAPTER 5 Polynomial and Rational Functions
SUMMARY
Analyzing the Graph of a Rational Function R
STEP 1: Factor the numerator and denominator of R. Find the domain of the rational function.
STEP 2: Write R in lowest terms.
STEP 3: Find and plot the intercepts of the graph. Use multiplicity to determine the behavior of the graph of R at
each x-intercept.
STEP 4: Find the vertical asymptotes. Graph each vertical asymptote using a dashed line. Determine the behavior of
the graph of R on either side of each vertical asymptote.
STEP 5: Find the horizontal or oblique asymptote, if one exists. Find points, if any, at which the graph of R intersects
this asymptote. Graph the asymptote using a dashed line. Plot any points at which the graph of R intersects
the asymptote.
STEP 6: Use the zeros of the numerator and denominator of R to divide the x-axis into intervals. Determine where
the graph of R is above or below the x-axis by choosing a number in each interval and evaluating R there.
Plot the points found.
STEP 7: Use the results obtained in Steps 1 through 6 to graph R.
Now Work
EX AMPLE 2
PROBLEM
7
Analyzing the Graph of a Rational Function
Analyze the graph of the rational function: R 1x2 =
Solution
STEP 4:
STEP 5:
x2 - 1
x2
1
1
=
= x x
x
x
x
1
Since S 0 as x S q , y = x is the
x
■
oblique asymptote.
1x + 12 1x - 12
. The domain of R is 5 x x ≠ 06 .
x
R is in lowest terms.
Because x cannot equal 0, there is no y-intercept. The graph has two
x-intercepts, - 1 and 1, each with odd multiplicity. Plot the points 1 - 1, 02
and 11, 02 . The graph will cross the x-axis at both points.
The real zero of the denominator with R in lowest terms is 0, so the graph of
R has the line x = 0 (the y-axis) as a vertical asymptote. Graph x = 0 using
a dashed line. The multiplicity of 0 is odd, so the graph will approach q on
one side of the asymptote x = 0, and - q on the other side.
Since the degree of the numerator, 2, is one greater than the degree of the
denominator, 1, the rational function will have an oblique asymptote. To
find the oblique asymptote, use long division.
STEP 1: R 1x2 =
STEP 2:
STEP 3:
NOTE Because the denominator of
the rational function is a monomial, we
can also find the oblique asymptote as
follows:
x2 - 1
x
x
x ) x2 - 1
x2
-1
The quotient is x, so the line y = x is an oblique asymptote of the graph.
Graph y = x using a dashed line.
To determine whether the graph of R intersects the asymptote y = x,
solve the equation R 1x2 = x.
R 1x2 =
x2 - 1
= x
x
x2 - 1 = x2
- 1 = 0 Impossible
x2 - 1
The equation
= x has no solution, so the graph of R does not intersect
x
the line y = x.
357
SECTION 5.3 The Graph of a Rational Function
STEP 6: The zeros of the numerator are - 1 and 1; the zero of the denominator is 0.
Use these values to divide the x-axis into four intervals:
1 - q , - 12
1 - 1, 02
10, 12
11, q 2
Now construct Table 10. Plot the points from Table 10. You should now have
Figure 32(a).
–1
Table 10
( - 1, 0)
(0, 1)
Number chosen
-2
1
2
1
2
2
Value of R
R( - 2) = -
3
1
Ra - b =
2
2
1
3
Ra b = 2
2
R(2) =
Location of graph
Below x-axis
Above x-axis
Below x-axis
Above x-axis
Point on graph
3
a - 2, - b
2
1 3
a- , b
2 2
1 3
a ,- b
2 2
3
a2, b
2
x=0
y
y
y=x
3
(− 1–2 , 3–2 )
(2, )
3
–
2
(−1, 0)
(1, 0)
)
( 1–2 , −3–2)
−3
(a)
3
x
(− 1–2 , 3–2 )
(2, )
3
–
2
(−2,
(2, 3–2 )
(−1, 0)
−3
−3–2
y=x
3
(−1, 0)
−3
3
2
x=0
y
y=x
(− 1–2 , 3–2 )
Figure 32
3
2
Since the graph of R is above the x-axis for - 1 6 x 6 0, the graph of
R will approach the vertical asymptote x = 0 at the top to the left of x = 0
[ lim-R 1x2 = q ]; since the graph of R approaches q on one side of the
xS 0
asymptote and - q on the other, the graph of R will approach the vertical
asymptote x = 0 at the bottom to the right of x = 0 [ lim+ R(x) = - q ].
xS 0
See Figure 32(b).
The complete graph is given in Figure 32(c).
x=0
3
x
(1, q )
STEP 7: The graph crosses the x-axis at x = −1 and x = 1, changing from being
below the x-axis to being above it in both cases.
Since the graph of R is below the x-axis for x 6 - 1 and is above the
x-axis for x 7 1, and since the graph of R does not intersect the oblique
asymptote y = x, the graph of R will approach the line y = x as shown in
Figure 32(b).
NOTE Notice that R in Example 2 is an
odd function. Do you see the symmetry
about the origin in the graph of R in
Figure 32(c)?
■
(−2,
1
( - q , - 1)
Interval
−3–2
0
(1, 0)
)
(1–2 , −3–2)
−3
(b)
3
x
−3
(−2,
−3–2
(1, 0)
)
3
x
( 1–2 , −3–2 )
−3
(c)
r
358
CHAPTER 5 Polynomial and Rational Functions
Seeing the Concept
x2 - 1
and compare what you see with Figure 32(c). Could you have predicted from the
x
graph that y = x is an oblique asymptote? Graph y = x and ZOOM-OUT. What do you observe?
Graph R(x) =
Now Work
PROBLEM
15
Analyzing the Graph of a Rational Function
EX AMPLE 3
Analyze the graph of the rational function: R(x) =
Solution
x0
y
6
y x2
STEP 1: R is completely factored. The domain of R is 5 x x ≠ 06 .
STEP 2: R is in lowest terms.
STEP 3: There is no y-intercept. Since x4 + 1 = 0 has no real solutions, there are no
x-intercepts.
STEP 4: R is in lowest terms, so x = 0 (the y-axis) is a vertical asymptote of R. Graph
the line x = 0 using dashes. The multiplicity of 0 is even, so the graph will
approach either q or - q on both sides of the asymptote.
STEP 5: Since the degree of the numerator, 4, is two more than the degree of the
denominator, 2, the rational function will not have a horizontal or oblique
asymptote. Find the end behavior of R. As x S q ,
R 1x2 =
(1, 2)
(1, 2)
3
3
x
(a)
x4 + 1
x2
x4 + 1
x4
≈
= x2
x2
x2
The graph of R will approach the graph of y = x2 as x S - q and as x S q .
The graph of R does not intersect y = x2. Do you know why? Graph y = x2
using dashes.
STEP 6: The numerator has no real zeros, and the denominator has one real zero at 0.
Divide the x-axis into the two intervals
1 - q , 02
x0
10, q 2
and construct Table 11.
y
6
y x2
0
Table 11
(1, 2)
(1, 2)
3
3
x
x
Interval
( - q , 0)
(0, q )
Number chosen
-1
1
Value of R
R( - 1) = 2
R(1) = 2
Location of graph
Above x-axis
Above x-axis
Point on graph
( - 1, 2)
(1, 2)
(b)
Figure 33
NOTE Notice that R in Example 3 is an
even function. Do you see the symmetry
about the y-axis in the graph of R?
■
STEP 7: Since the graph of R is above the x-axis and does not intersect y = x2, place
arrows above y = x2 as shown in Figure 33(a). Also, since the graph of R is
above the x-axis, and the multiplicity of the zero that gives rise to the vertical
asymptote, x = 0, is even, it will approach the vertical asymptote x = 0 at the
top to the left of x = 0 and at the top to the right of x = 0. See Figure 33(a).
Figure 33(b) shows the complete graph.
r
SECTION 5.3 The Graph of a Rational Function
359
Seeing the Concept
x4 + 1
and compare what you see with Figure 33(b). Use MINIMUM to find the two turning
x2
2
points. Enter Y2 = x and ZOOM-OUT. What do you see?
Graph R(x) =
Now Work
EXAMPL E 4
PROBLEM
13
Analyzing the Graph of a Rational Function
Analyze the graph of the rational function: R 1x2 =
Solution
3x2 - 3x
x + x - 12
2
STEP 1: Factor R to get
R 1x2 =
STEP 2:
STEP 3:
STEP 4:
STEP 5:
3x1x - 12
1x + 42 1x - 32
The domain of R is 5 x x ≠ - 4, x ≠ 36 .
R is in lowest terms.
The y-intercept is R 102 = 0. Plot the point 10, 02 . Since the real solutions
of the equation 3x1x - 12 = 0 are x = 0 and x = 1, the graph has two
x-intercepts, 0 and 1, each with odd multiplicity. Plot the points 10, 02 and
11, 02; the graph will cross the x-axis at both points.
R is in lowest terms. The real solutions of the equation 1x + 42 1x - 32 = 0
are x = - 4 and x = 3, so the graph of R has two vertical asymptotes, the
lines x = - 4 and x = 3. Graph these lines using dashes. The multiplicities
that give rise to the vertical asymptotes are both odd, so the graph will approach
q on one side of each vertical asymptote and - q on the other side.
Since the degree of the numerator equals the degree of the denominator,
the graph has a horizontal asymptote. To find it, form the quotient of the
leading coefficient of the numerator, 3, and the leading coefficient of the
denominator, 1. The graph of R has the horizontal asymptote y = 3.
To find out whether the graph of R intersects the asymptote, solve the
equation R 1x2 = 3.
R 1x2 =
3x2 - 3x
= 3
x + x - 12
3x2 - 3x = 3x2 + 3x - 36
- 6x = - 36
x = 6
2
The graph intersects the line y = 3 at x = 6, and 16, 32 is a point on the
graph of R. Plot the point 16, 32 and graph the line y = 3 using dashes.
STEP 6: The real zeros of the numerator, 0 and 1, and the real zeros of the denominator,
- 4 and 3, divide the x-axis into five intervals:
1 - q , - 42
1 - 4, 02
10, 12
11, 32
13, q 2
Construct Table 12. See page 360. Plot the points from Table 12. Figure 34(a)
shows the graph so far.
STEP 7: Since the graph of R is above the x-axis for x 6 - 4 and only crosses the
line y = 3 at 16, 32 , as x approaches - q the graph of R will approach
the horizontal asymptote y = 3 from above 1 lim R 1x2 = 32. The graph
xS -q
of R will approach the vertical asymptote x = - 4 at the top to the left
of x = - 4 1 lim -R 1x2 = q 2 and at the bottom to the right of x = - 4
x S -4
1 lim +R 1x2 = - q 2 . The graph of R will approach the vertical asymptote
x S -4
360
CHAPTER 5 Polynomial and Rational Functions
Table 12
–4
0
( - q , - 4)
Interval
1
3
x
( - 4, 0)
(0, 1)
(1, 3)
(3, q )
2
4
Number chosen
-5
-2
1
2
Value of R
R( - 5) = 11.25
R( - 2) = - 1.8
1
1
Ra b =
2
15
R(2) = - 1
R(4) = 4.5
Location of graph
Above x-axis
Below x-axis
Above x-axis
Below x-axis
Above x-axis
Point on graph
( - 5, 11.25)
( - 2, - 1.8)
1 1
a , b
2 15
(2, - 1)
(4, 4.5)
x = 3 at the bottom to the left of x = 3 1 lim-R 1x2 = - q 2 and at the top to
xS3
the right of x = 3 1 lim+R 1x2 = q 2 .
S
x 3
We do not know whether the graph of R crosses or touches the line y = 3 at
16, 32. To see whether the graph, in fact, crosses or touches the line y = 3, plot
63
an additional point to the right of 16, 32. We use x = 7 to find R 172 =
6 3.
22
The graph crosses y = 3 at x = 6. Because 16, 32 is the only point where the
graph of R intersects the asymptote y = 3, the graph must approach the line
y = 3 from below as x S q 1 lim R 1x2 = 32 .
x Sq
The graph crosses the x-axis at x = 0, changing from being below the x-axis
to being above. The graph also crosses the x-axis at x = 1, changing from being
above the x-axis to being below. See Figure 34(b). The complete graph is shown
in Figure 34(c).
x = −4
(−5, 11.25)
y
x 4
x=3
(5, 11.25)
10
1
( 1–2 , ––
15)
(4, 4.5)
(6, 3)
(0, 0)
−5
(−2, −1.8)
y
(2, −1)
10
1
( 1–2 , ––
15)
y=3
(4, 4.5)
(6, 3)
(0, 0)
5
x
5
x3
(2, 1.8)
(2, 1)
5
y3
x
–– )
(7, 63
22
(1, 0)
(1, 0)
−10
– 10
(b)
(a)
x 4
(5, 11.25)
y
x3
10
1
( 1–2 , ––
15)
(4, 4.5)
(6, 3)
(0, 0)
5
(2, 1.8)
(2, 1)
5
y3
x
–– )
(7, 63
22
(1, 0)
– 10
Figure 34
(c)
r
SECTION 5.3 The Graph of a Rational Function
361
Exploration
3x2 - 3x
x + x - 12
Result Figure 35(a) shows the graph. The graph does not clearly display the behavior of the function
between the two x-intercepts, 0 and 1. Nor does it clearly display the fact that the graph crosses the
horizontal asymptote at (6, 3). To see these parts better, graph R for - 1 … x … 2 [Figure 35(b)] and for
4 … x … 60 [Figure 35(c)].
Graph the rational function: R(x) =
2
3.5
0.5
10
21
210
2
y5 3
10
Figure 35
4
21
210
(a)
60
2.5
(b)
(c)
The new graphs reflect the behavior produced by the analysis. Furthermore, we observe two turning
points, one between 0 and 1 and the other to the right of 6. Rounded to two decimal places, these
turning points are (0.52, 0.07) and (11.48, 2.75).
Now Work
EXAMPL E 5
PROBLEM
31
Analyzing the Graph of a Rational Function with a Hole
Analyze the graph of the rational function: R 1x2 =
Solution
2x2 - 5x + 2
x2 - 4
STEP 1: Factor R and obtain
R 1x2 =
12x - 12 1x - 22
1x + 22 1x - 22
The domain of R is 5 x x ≠ - 2, x ≠ 26 .
STEP 2: In lowest terms,
R 1x2 =
2x - 1
x + 2
x ≠ - 2, x ≠ 2
1
1
STEP 3: The y-intercept is R 102 = - . Plot the point a0, - b . The graph has one
2
2
1
1
x-intercept, , with odd multiplicity. Plot the point ¢ , 0≤. The graph will
2
2
1
cross the x-axis at x = . See Figure 36(a) on page 363.
2
STEP 4: Since x + 2 is the only factor of the denominator of R 1x2 in lowest terms,
the graph has one vertical asymptote, x = - 2. However, the rational
function is undefined at both x = 2 and x = - 2. Graph the line x = - 2
using dashes. The multiplicity of - 2 is odd, so the graph will approach q on
one side of the vertical asymptote and - q on the other side.
STEP 5: Since the degree of the numerator equals the degree of the denominator,
the graph has a horizontal asymptote. To find it, form the quotient of the
leading coefficient of the numerator, 2, and the leading coefficient of the
denominator, 1. The graph of R has the horizontal asymptote y = 2. Graph
the line y = 2 using dashes.
362
CHAPTER 5 Polynomial and Rational Functions
To find out whether the graph of R intersects the horizontal asymptote
y = 2, solve the equation R 1x2 = 2.
R 1x2 =
2x - 1
x + 2
2x - 1
2x - 1
-1
= 2
= 21x + 22
= 2x + 4
= 4
Impossible
The graph does not intersect the line y = 2.
STEP 6: Look at the factored expression for R in Step 1. The real zeros of the numerator
1
and denominator, - 2, , and 2, divide the x-axis into four intervals:
2
1 - q , - 22
1
a - 2, b
2
1
a , 2b
2
12, q 2
Construct Table 13. Plot the points in Table 13.
–2
Table 13
1/2
2
x
Interval
( - q , - 2)
1
a - 2, b
2
1
a , 2b
2
(2, q )
Number chosen
-3
-1
1
3
R(3) = 1
Value of R
R( - 3) = 7
R( - 1) = - 3
1
R(1) =
3
Location of graph
Above x-axis
Below x-axis
Above x-axis
Above x-axis
Point on graph
( - 3, 7)
( - 1, - 3)
1
a1, b
3
(3, 1)
NOTE The coordinates of the hole were
obtained by evaluating R in lowest terms
2x - 1
,
at x = 2. R in lowest terms is
x + 2
2(2) - 1
3
which, at x = 2, is
= . ■
2 + 2
4
STEP 7: From Table 13 we know that the graph of R is above the x-axis for x 6 - 2.
From Step 5 we know that the graph of R does not intersect the
asymptote y = 2. Therefore, the graph of R will approach y = 2 from
above as x S - q and will approach the vertical asymptote x = - 2 at the
top from the left.
1
Since the graph of R is below the x-axis for - 2 6 x 6 , the graph will
2
approach x = - 2 at the bottom from the right. Finally, since the graph of
1
R is above the x-axis for x 7 and does not intersect the horizontal asymptote
2
y = 2, the graph of R will approach y = 2 from below as x S q . The graph
1
crosses the x-axis at x = , changing from being below the x-axis to being
2
above. See Figure 36(a).
See Figure 36(b) for the complete graph. Since R is not defined at 2,
3
there is a hole at the point a2, b .
4
Exploration
2x2 - 5x + 2
3
. Do you see the hole at a2, b? TRACE along the graph. Did you obtain an
4
x2 - 4
ERROR at x = 2? Are you convinced that an algebraic analysis of a rational function is required in order
to accurately interpret the graph obtained with a graphing utility?
Graph R(x) =
SECTION 5.3 The Graph of a Rational Function
y
x 2
y
x 2
8
(3, 7)
(3, 7)
6
4
(1, )
(0,
4
3
2
1–
3
2
)
1
(
y2
(3, 1)
1
2
8
6
4
1–2
1–
,
2
2
3
(0, 1–2 )
4
x
3
2
0)
Hole at
3
(1, ) (2, –4 )
1–
3
2
1
2
(1, 3)
(a)
(b)
(
y2
(3, 1)
1
(1, 3)
Figure 36
363
1–
,
2
2
3
x
0)
r
As Example 5 shows, the zeros of the denominator of a rational function give
rise to either vertical asymptotes or holes on the graph.
Now Work
EXAMPL E 6
PROBLEM
33
Constructing a Rational Function from Its Graph
Find a rational function that might have the graph shown in Figure 37.
x 5
y
x2
10
5
y2
15
10
5
5
10
15 x
5
10
Figure 37
Solution
p 1x2
in lowest terms determines the
q 1x2
x-intercepts of its graph. The graph shown in Figure 37 has x-intercepts - 2 (even
multiplicity; graph touches the x-axis) and 5 (odd multiplicity; graph crosses the
x-axis). So one possibility for the numerator is p 1x2 = 1x + 22 2 1x - 52.
The denominator of a rational function in lowest terms determines the vertical
asymptotes of its graph. The vertical asymptotes of the graph are x = - 5 and x = 2.
Since R 1x2 approaches q to the left of x = - 5 and R 1x2 approaches - q to
the right of x = - 5, we know that (x + 5) is a factor of odd multiplicity in q 1x2 .
Also, R 1x2 approaches - q on both sides of x = 2, so (x - 2) is a factor of even
multiplicity in q 1x2 . A possibility for the denominator is q 1x2 = 1x + 52 1x - 22 2.
1x + 22 2 1x - 52
So far we have R 1x2 =
.
1x + 52 1x - 22 2
The numerator of a rational function R 1x2 =
364
CHAPTER 5 Polynomial and Rational Functions
However, the horizontal asymptote of the graph given in Figure 37 is y = 2, so
we know that the degree of the numerator must equal the degree of the denominator
2
and that the quotient of leading coefficients must be . This leads to
1
5
215
21x + 22 2 1x - 52
10
R 1x2 =
1x + 52 1x - 22 2
25
r
Check: Figure 38 shows the graph of R on a graphing utility. Since Figure 38
looks similar to Figure 37, we have found a rational function R for the
graph in Figure 37.
Figure 38
Now Work
PROBLEM
51
2 Solve Applied Problems Involving Rational Functions
EX AMPLE 7
Finding the Least Cost of a Can
Reynolds Metal Company manufactures aluminum cans in the shape of a cylinder
1
with a capacity of 500 cubic centimeters a literb . The top and bottom of the can are
2
made of a special aluminum alloy that costs 0.05. per square centimeter. The sides of
the can are made of material that costs 0.02. per square centimeter.
(a)
(b)
(c)
(d)
Solution
Top
r
Area ⫽ ␲r 2
r
h h
Lateral Surface
Area ⫽ 2␲rh
Express the cost of material for the can as a function of the radius r of the can.
Use a graphing utility to graph the function C = C 1r2.
What value of r will result in the least cost?
What is this least cost?
(a) Figure 39 illustrates the components of a can in the shape of a right circular
cylinder. Notice that the material required to produce a cylindrical can of height h
and radius r consists of a rectangle of area 2prh and two circles, each of area pr 2.
The total cost C (in cents) of manufacturing the can is therefore
C = Cost of the top and bottom + Cost of the side
= 21pr 2 2 # 10.052
+ 12prh2 # 10.022
3
Area ⫽ ␲r 2
Total area
of top and
bottom
3
3
3
Cost/unit
area
Total
area of
side
Cost/unit
area
= 0.10pr 2 + 0.04prh
Bottom
Figure 39
There is an additional restriction that the height h and radius r must be chosen so
that the volume V of the can is 500 cubic centimeters. Since V = pr 2h, we have
500 = pr 2h so h =
500
pr 2
Substituting this expression for h, we find that the cost C, in cents, as a function
of the radius r is
C 1r2 = 0.10pr 2 + 0.04pr
60
0
0
Figure 40
10
#
500
0.10pr 3 + 20
20
=
= 0.10pr 2 +
2
r
r
pr
(b) See Figure 40 for the graph of C = C 1r2.
(c) Using the MINIMUM command, the cost is least for a radius of about
3.17 centimeters.
(d) The least cost is C 13.172 ≈ 9.47..
r
Now Work
PROBLEM
61
SECTION 5.3 The Graph of a Rational Function
365
5.3 Assess Your Understanding
‘Are You Prepared?’ The answer is given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Find the intercepts of the graph of the equation y =
x2 - 1
. (pp. 159–160)
x2 - 4
Concepts and Vocabulary
2. True or False Every rational function has at least one
asymptote.
3. Which type of asymptote will never intersect the graph of a
rational function?
(a) horizontal (b) oblique (c) vertical (d) all of these
4. Identify the y-intercept of the graph of
61x - 12
R 1x2 =
.
1x + 12 1x + 22
(a) - 3
(b) - 2
(c) - 1
5. R 1x2 =
x1x - 22 2
x - 2
(a) Find the domain of R.
(b) Find the x-intercepts of R.
6. True or False The graph of a rational function sometimes
has a hole.
(d) 1
Skill Building
In Problems 7–50, follow Steps 1 through 7 on page 356 to analyze the graph of each function.
7. R 1x2 =
x + 1
x1x + 42
8. R 1x2 =
x
1x - 12 1x + 22
9. R 1x2 =
3x + 3
2x + 4
10. R 1x2 =
2x + 4
x - 1
11. R 1x2 =
3
x - 4
12. R 1x2 =
6
x - x - 6
13. P 1x2 =
x4 + x2 + 1
x2 - 1
14. Q1x2 =
x4 - 1
x2 - 4
15. H1x2 =
x3 - 1
x2 - 9
16. G1x2 =
x3 + 1
x2 + 2x
17. R 1x2 =
x2
x2 + x - 6
18. R 1x2 =
x2 + x - 12
x2 - 4
19. G1x2 =
x
x - 4
20. G1x2 =
3x
x - 1
21. R 1x2 =
3
1x - 12 1x2 - 42
22. R 1x2 =
-4
1x + 12 1x2 - 92
23. H1x2 =
x2 - 1
x4 - 16
24. H1x2 =
x2 + 4
x4 - 1
25. F 1x2 =
x2 - 3x - 4
x + 2
26. F 1x2 =
x2 + 3x + 2
x - 1
27. R 1x2 =
x2 + x - 12
x - 4
28. R 1x2 =
x2 - x - 12
x + 5
29. F 1x2 =
x2 + x - 12
x + 2
30. G1x2 =
x2 - x - 12
x + 1
31. R 1x2 =
x1x - 12 2
34. R 1x2 =
x2 + 3x - 10
x2 + 8x + 15
35. R 1x2 =
37. R 1x2 =
x2 + 5x + 6
x + 3
38. R 1x2 =
40. H1x2 =
2 - 2x
x2 - 1
41. F 1x2 =
44. G1x2 =
2 - x
1x - 12 2
45. f 1x2 = x +
1
x
46. f 1x2 = 2x +
9
x
49. f 1x2 = x +
1
x3
50. f 1x2 = 2x +
9
x3
2
2
16
x
2
32. R 1x2 =
1x + 32 3
48. f 1x2 = 2x2 +
2
x2 - 5x + 4
x2 - 2x + 1
1x - 12 1x + 22 1x - 32
33. R 1x2 =
x2 + x - 12
x2 - x - 6
6x2 - 7x - 3
2x2 - 7x + 6
36. R 1x2 =
8x2 + 26x + 15
2x2 - x - 15
x2 + x - 30
x + 6
39. H1x2 =
3x - 6
4 - x2
x1x - 42 2
42. F 1x2 =
x2 - 2x - 15
x2 + 6x + 9
43. G1x2 =
x
1x + 22 2
47. f 1x2 = x2 +
1
x
366
CHAPTER 5 Polynomial and Rational Functions
In Problems 51–54, find a rational function that might have the given graph. (More than one answer might be possible.)
51.
x 2
52.
x2
y
x 1
y
3
x1
3
y1
y0
3
3 x
3
3
53.
x
3
3
54.
y
x 3 y
10
3
x4
8
2
y1
6
4
4 3 2
1
1
3
4
5
x
2
x 1
y3
2
15
10
5
x2
5
10
15
20 x
2
4
6
8
Applications and Extensions
55. Drug Concentration The concentration C of a certain drug
in a patient’s bloodstream t hours after injection is given by
C 1t2 =
t
2t 2 + 1
(a) Find the horizontal asymptote of C 1t2 . What
happens to the concentration of the drug as t increases?
(b) Using your graphing utility, graph C = C 1t2.
(c) Determine the time at which the concentration is
highest.
56. Drug Concentration The concentration C of a certain drug
in a patient’s bloodstream t minutes after injection is given by
50t
C 1t2 = 2
t + 25
(a) Find the horizontal asymptote of C 1t2 . What happens
to the concentration of the drug as t increases?
(b) Using your graphing utility, graph C = C 1t2.
(c) Determine the time at which the concentration is
highest.
57. Minimum Cost A rectangular area adjacent to a river is
to be fenced in; no fence is needed on the river side. The
enclosed area is to be 1000 square feet. Fencing for the side
parallel to the river is $5 per linear foot, and fencing for
the other two sides is $8 per linear foot; the four corner posts
are $25 apiece. Let x be the length of one of the sides
perpendicular to the river.
(a) Write a function C 1x2 that describes the cost of the project.
(b) What is the domain of C?
(c) Use a graphing utility to graph C = C 1x2.
(d) Find the dimensions of the cheapest enclosure.
Source: http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/
pandr/rational/rational.html
58. Doppler Effect The Doppler effect (named after Christian
Doppler) is the change in the pitch (frequency) of the sound
from a source (s) as heard by an observer (o) when one or
both are in motion. If we assume both the source and the
observer are moving in the same direction, the relationship is
f′ = fa a
where
f′
fa
v
vo
vs
=
=
=
=
=
v - vo
b
v - vs
perceived pitch by the observer
actual pitch of the source
speed of sound in air (assume 772.4 mph)
speed of the observer
speed of the source
Suppose that you are traveling down the road at 45 mph
and you hear an ambulance (with siren) coming toward you
from the rear. The actual pitch of the siren is 600 hertz (Hz).
(a) Write a function f′(vs) that describes this scenario.
(b) If f′ = 620 Hz, find the speed of the ambulance.
(c) Use a graphing utility to graph the function.
(d) Verify your answer from part (b).
Source: www.acs.psu.edu/drussell/
SECTION 5.3 The Graph of a Rational Function
59. Minimizing Surface Area United Parcel Service has
contracted you to design a closed box with a square base
that has a volume of 10,000 cubic inches. See the illustration.
(a) Express the surface area S of the box
as a function of x.
y
(b) Using a graphing utility, graph the
function found in part (a).
x
(c) What is the minimum amount of
x
cardboard that can be used to construct
the box?
(d) What are the dimensions of the box that minimize the
surface area?
(e) Why might UPS be interested in designing a box that
minimizes the surface area?
60. Minimizing Surface Area United Parcel
y
Service has contracted you to design an open
box with a square base that has a volume of
x
5000 cubic inches. See the illustration.
x
(a) Express the surface area S of the box as
a function of x.
(b) Using a graphing utility, graph the function found in part (a).
(c) What is the minimum amount of cardboard that can be
used to construct the box?
(d) What are the dimensions of the box that minimize the
surface area?
(e) Why might UPS be interested in designing a box that
minimizes the surface area?
61. Cost of a Can A can in the shape of a right circular cylinder is
required to have a volume of 500 cubic centimeters. The top
367
and bottom are made of material that costs 6¢ per square
centimeter, while the sides are made of material that costs
4¢ per square centimeter.
(a) Express the total cost C of the material as a function of
the radius r of the cylinder. (Refer to Figure 39.)
(b) Graph C = C 1r2. For what value of r is the cost C a
minimum?
62. Material Needed to Make a Drum A steel drum in the shape
of a right circular cylinder is required to have a volume of
100 cubic feet.
(a) Express the amount A of material required to make the
drum as a function of the radius r of the cylinder.
(b) How much material is required if the drum’s radius
is 3 feet?
(c) How much material is required if the drum’s radius
is 4 feet?
(d) How much material is required if the drum’s radius
is 5 feet?
(e) Graph A = A(r). For what value of r is A smallest?
Discussion and Writing
at - 1; one vertical asymptote at x = - 5 and another at
x = 6; and one horizontal asymptote, y = 3. Compare
your function to a fellow classmate’s. How do they differ?
What are their similarities?
63. Graph each of the following functions:
2
3
y =
x - 1
x - 1
y =
x - 1
x - 1
y =
x4 - 1
x - 1
y =
x5 - 1
x - 1
Is x = 1 a vertical asymptote? Why not? What is happening for
xn - 1
,n Ú 1
x = 1? What do you conjecture about y =
x - 1
an integer, for x = 1?
67. Create a rational function that has the following characteristics:
crosses the x-axis at 3; touches the x-axis at - 2; one vertical
asymptote, x = 1; and one horizontal asymptote, y = 2.
Give your rational function to a fellow classmate and ask
for a written critique of your rational function.
65. Write a few paragraphs that provide a general strategy for
graphing a rational function. Be sure to mention the following:
proper, improper, intercepts, and asymptotes.
68. Create a rational function with the following characteristics:
three real zeros, one of multiplicity 2; y-intercept 1; vertical
asymptotes, x = - 2 and x = 3; oblique asymptote,
y = 2x + 1. Is this rational function unique? Compare your
function with those of other students. What will be the same
as everyone else’s? Add some more characteristics, such as
symmetry or naming the real zeros. How does this modify
the rational function?
66. Create a rational function that has the following
characteristics: crosses the x-axis at 2; touches the x-axis
69. Explain the circumstances under which the graph of a
rational function will have a hole.
64. Graph each of the following functions:
y =
x2
x - 1
y =
x4
x - 1
y =
x6
x - 1
y =
x8
x - 1
What similarities do you see? What differences?
Retain Your Knowledge
Problems 70–73 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
70. Subtract: (4x3 - 7x + 1) - (5x2 - 9x + 3)
71. Solve:
x - 2
3x
=
3x + 1
x + 5
2
72. Find the maximum value of f 1x2 = - x2 + 6x - 5.
3
25 - 3
. Round your answer to three
73. Approximate
27 + 2
decimal places.
368
CHAPTER 5 Polynomial and Rational Functions
‘Are You Prepared?’ Answers
1
1. ¢0, ≤, 11, 02, 1 - 1, 02
4
5.4 Polynomial and Rational Inequalities
PREPARING FOR THIS SECTION Before getting started, review the following:
r Solving Linear Inequalities (Section 1.5,
pp. 123–125)
r Solving Quadratic Inequalities (Section 4.5,
pp. 312–313)
Now Work the ‘Are You Prepared?’ problems on page 372.
OBJECTIVES 1 Solve Polynomial Inequalities (p. 368)
2 Solve Rational Inequalities (p. 370)
1 Solve Polynomial Inequalities
In this section we solve inequalities that involve polynomials of degree 3 and higher,
along with inequalities that involve rational functions. To help understand the
algebraic procedure for solving such inequalities, we use the information obtained
in the previous three sections about the graphs of polynomial and rational functions.
The approach follows the same methodology that we used to solve inequalities
involving quadratic functions.
EX AMPLE 1
Solution
Solving a Polynomial Inequality Using Its Graph
Solve 1x + 32 1x - 12 2 7 0 by graphing f1x2 = 1x + 32 1x - 12 2.
Graph f 1x2 = 1x + 32 1x - 12 2 and determine the intervals of x for which the
graph is above the x-axis. These values of x result in f 1x2 being positive. Using
Steps 1 through 5 on page 335, we obtain the graph shown in Figure 41.
y
12
End behavior
y = x3
(22, 9)
9
6
3
(23, 0)
24
22
3
End behavior
3
y=x
2
(1, 0)
4
x
6
Figure 41 f (x) = (x + 3)(x - 1)2
From the graph, we can see that f 1x2 7 0 for - 3 6 x 6 1 or x 7 1. The solution
set is 5 x - 3 6 x 6 1 or x 7 16 or, using interval notation, 1 - 3, 12 h 11, q 2 .
r
Now Work
PROBLEM
9
The results of Example 1 lead to the following approach to solving polynomial
and rational inequalities algebraically. Suppose that the polynomial or rational
inequality is in one of the forms
f1x2 6 0
f1x2 7 0
f1x2 … 0
f1x2 Ú 0
369
SECTION 5.4 Polynomial and Rational Inequalities
Locate the zeros of f if f is a polynomial function, and locate the zeros of the
numerator and the denominator if f is a rational function. If we use these zeros to
divide the real number line into intervals, we know that on each interval, the graph
of f is either above the x-axis 3 f1x2 7 04 or below the x-axis 3 f 1x2 6 04 . This will
enable us to identify the solution of the inequality.
EXAMPL E 2
How to Solve a Polynomial Inequality Algebraically
Solve the inequality x4 7 x algebraically, and graph the solution set.
Step-by-Step Solution
Rearrange the inequality so that 0 is on the right side.
Step 1: Write the inequality so
that a polynomial expression f is
on the left side and zero is on the
right side.
x4 7 x
x4 - x 7 0
Subtract x from both sides of the inequality.
This inequality is equivalent to the one we wish to solve.
Step 2: Determine the real zeros
(x-intercepts of the graph) of f.
Find the real zeros of f1x2 = x4 - x by solving x4 - x = 0.
x = 0 or
x = 0 or
x4 - x
x1x3 - 12
2
x1x - 12 1x + x + 12
x - 1 = 0 or x2 + x + 1
x = 1
=
=
=
=
0
0 Factor out x.
0 Factor the difference of two cubes.
0 Set each factor equal to zero and solve.
The equation x2 + x + 1 = 0 has no real solutions. Do you see why?
Step 3: Use the zeros found in
Step 2 to divide the real number line
into intervals.
Use the real zeros to separate the real number line into three intervals:
Step 4: Select a number in each
interval, evaluate f at the number,
and determine whether f (x) is positive or negative. If f (x) is positive, all
values of f in the interval are positive. If f (x) is negative, all values of f
in the interval are negative.
Select a test number in each interval found in Step 3 and evaluate f1x2 = x4 - x at
each number to determine whether f 1x2 is positive or negative. See Table 14.
NOTE If the inequality is not strict (that
is, if it is … or Ú ), include the solutions of
f (x) = 0 in the solution set.
■
–2 –1 0 1 2
Figure 42
x
1 - q , 02
10, 12
Table 14
11, q 2
0
1
x
Interval
( - q , 0)
(0, 1)
(1, q )
Number chosen
-1
1
2
2
Value of f
f ( - 1) = 2
1
7
fa b = 2
16
f (2) = 14
Conclusion
Positive
Negative
Positive
Since we want to know where f1x2 is positive, conclude that f1x2 7 0 for all
numbers x for which x 6 0 or x 7 1. Because the original inequality is strict, numbers x
that satisfy the equation x4 = x are not solutions. The solution set of the inequality
x4 7 x is {x x 6 0 or x 7 1} or, using interval notation, 1 - q , 02 h 11, q 2 .
Figure 42 shows the graph of the solution set.
r
The Role of Multiplicity in Solving Polynomial Inequalities
In Example 2, we used the number - 1 and found that f is positive for all x 6 0.
Because the “cut point” of 0 is the result of a zero of odd multiplicity (x is a factor
to the first power), we know that the sign of f will change on either side of 0, so
370
CHAPTER 5 Polynomial and Rational Functions
for 0 6 x 6 1, f(x) will be negative. Similarly, we know that f(x) will be positive
for x 7 1, since the multiplicity of the zero 1 is odd. Therefore, the solution set of
x4 7 x is {x x 6 0 or x 7 16 or, using interval notation, 1 - q , 02 h 11, q 2.
Now Work
PROBLEM
21
2 Solve Rational Inequalities
Just as we presented a graphical approach to help us understand the algebraic
procedure for solving inequalities involving polynomials, we present a graphical
approach to help us understand the algebraic procedure for solving inequalities
involving rational expressions.
EX AMPLE 3
Solving a Rational Inequality Using Its Graph
Solve
Solution
x - 1
x - 1
Ú 0 by graphing R 1x2 = 2
.
x2 - 4
x - 4
Graph R 1x2 =
x - 1
and determine the intervals of x such that the graph is above
x2 - 4
or on the x-axis. These values of x result in R 1x2 being positive or zero. We graphed
x - 1
R 1x2 = 2
in Example 1, Section 5.3 (pp. 353–355). We reproduce the graph in
x - 4
Figure 43.
x
2
x
y
2
3
(1, 0)
3
( 3,
0.8)
(0, 1–4)
(3–2, 2–7)
(3, 0.4)
3
x y
0
3
Figure 43 R 1x2 =
x - 1
x2 - 4
From the graph, we can see that R 1x2 Ú 0 for - 2 6 x … 1 or x 7 2. The solution
set is 5 x - 2 6 x … 1 or x 7 26 or, using interval notation, 1 - 2, 14 h 12, q 2 .
r
Now Work
PROBLEM
15
To solve a rational inequality algebraically, we follow the same approach that
we used to solve a polynomial inequality algebraically. However, we must also
identify the zeros of the denominator of the rational function, because the sign of
a rational function may change on either side of a vertical asymptote. Convince
yourself of this by looking at Figure 43. Notice that the function values are negative
for x 6 - 2 and are positive for x 7 - 2 (but less than 1).
EX AMPLE 4
How to Solve a Rational Inequality Algebraically
Solve the inequality
3x2 + 13x + 9
… 3 algebraically, and graph the solution set.
1x + 22 2
SECTION 5.4 Polynomial and Rational Inequalities
Step-by-Step Solution
371
Rearrange the inequality so that 0 is on the right side.
3x2 + 13x + 9
… 3
1x + 22 2
Step 1: Write the inequality so that
a rational expression f is on the left
side and zero is on the right side.
3x2 + 13x + 9
- 3 … 0
x2 + 4x + 4
Subtract 3 from both sides of the inequality;
Expand (x + 2)2.
3x2 + 13x + 9
x2 + 4x + 4
#
3
… 0
x2 + 4x + 4
x2 + 4x + 4
Multiply 3 by
3x2 + 13x + 9 - 3x2 - 12x - 12
… 0
x2 + 4x + 4
Write as a single quotient.
x - 3
… 0
1x + 22 2
x 2 + 4x + 4
.
x 2 + 4x + 4
Combine like terms.
x - 3
is 3. Also, f is undefined for x = - 2.
1x + 22 2
Step 2: Determine the real zeros
(x-intercepts of the graph) of f and
the real numbers for which f is
undefined.
The zero of f1x2 =
Step 3: Use the zeros and undefined
values found in Step 2 to divide the
real number line into intervals.
Use the zero and the undefined value to separate the real number line into three
intervals:
Step 4: Select a number in each
interval, evaluate f at the number,
and determine whether f (x) is positive
or negative. If f (x) is positive, all
values of f in the interval are positive.
If f (x) is negative, all values of f in
the interval are negative.
Select a test number in each interval from Step 3, and evaluate f at each number to
determine whether f 1x 2 is positive or negative. See Table 15.
NOTE If the inequality is not strict
( … or Ú ), include the solutions of
f(x) = 0 in the solution set.
■
6
4
Figure 44
2
0
2
4
6
x
1 - q , - 22
1 - 2, 32
13, q 2
–2
Table 15
3
x
Interval
( - q , - 2)
( - 2, 3)
(3, q )
Number chosen
-3
0
4
Value of f
f( - 3) = - 6
f (0) = -
Conclusion
Negative
Negative
3
4
f (4) =
1
36
Positive
Since we want to know where f(x) is negative or zero, we conclude that
f (x) … 0 for all numbers for which x 6 - 2 or - 2 6 x … 3. Notice that we do not
include - 2 in the solution because - 2 is not in the domain of f. The solution set of the
3x2 + 13x + 9
inequality
… 3 is {x x 6 - 2 or - 2 6 x … 3} or, using interval
1x + 22 2
notation, 1 - q , - 22 h 1 - 2, 3]. Figure 44 shows the graph of the solution set.
r
The Role of Multiplicity in Solving Rational Inequalities
In Example 4, we used the number - 3 and found that f(x) is negative for all x 6 - 2.
Because the “cut point” of - 2 is the result of a zero of even multiplicity, we know
the sign of f(x) will not change on either side of - 2, so for - 2 6 x 6 3, f(x) will also
be negative. Because the “cut point” of 3 is the result of a zero of odd multiplicity,
the sign of f(x) will change on either side of 3, so for x 7 3, f(x) will be positive.
3x2 + 13x + 9
Therefore, the solution set of
… 3 is {x x 6 - 2 or - 2 6 x … 3} or,
1x + 22 2
using interval notation, 1 - q , - 22 h 1 - 2, 3].
Now Work
PROBLEMS
33 and 39
372
CHAPTER 5 Polynomial and Rational Functions
SUMMARY
Steps for Solving Polynomial and Rational Inequalities Algebraically
STEP 1: Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right
side in one of the following forms:
f 1x2 7 0
f1x2 Ú 0
f1x2 6 0
f1x2 … 0
For rational expressions, be sure that the left side is written as a single quotient, and find the domain of f .
STEP 2: Determine the real numbers at which the expression f equals zero and, if the expression is rational, the
real numbers at which the expression f is undefined.
STEP 3: Use the numbers found in Step 2 to separate the real number line into intervals.
STEP 4: Select a number in each interval and evaluate f at the number.
(a) If the value of f is positive, then f1x2 7 0 for all numbers x in the interval.
(b) If the value of f is negative, then f1x2 6 0 for all numbers x in the interval.
If the inequality is not strict 1 Ú or … 2 , include the solutions of f1x2 = 0 that are in the domain of f in the
solution set. Be careful to exclude values of x where f is undefined.
5.4 Assess Your Understanding
‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Solve the inequality 3 - 4x 7 5. Graph the solution set.
(pp. 123–125)
2. Solve the inequality x2 - 5x … 24. Graph the solution set.
(pp. 312–313)
Concepts and Vocabulary
3. Which of the following could be a test number for the interval
- 2 6 x 6 5?
(a) - 3 (b) - 2 (c) 4 (d) 7
4. True or False The graph of f 1x2 =
x
is above the x-axis
x - 3
for x 6 0 or x 7 3, so the solution set of the inequality
x
Ú 0 is 5x x … 0 or x Ú 36 .
x - 3
Skill Building
In Problems 5–8, use the graph of the function f to solve the inequality.
5. (a) f 1x2 7 0
(b) f 1x2 … 0
6. (a) f 1x2 6 0
(b) f 1x2 Ú 0
y
2
y
1
0
1
2
–2
x
–1
1
–1
–2
7. (a) f 1x2 6 0
(b) f 1x2 Ú 0
x 1
y
8. (a) f 1x2 7 0
(b) f 1x2 … 0
x1
3
y
3
2
y0
3
3
4 3 2
3
y1
x
1
1
3
2
x 1
x2
4
5
x
2
3
x
SECTION 5.4 Polynomial and Rational Inequalities
373
In Problems 9–14, solve the inequality by using the graph of the function.
[Hint: The graphs were drawn in Problems 81–86 of Section 5.1.]
9. Solve f 1x2 6 0, where f 1x2 = x2 1x - 32 .
10. Solve f 1x2 … 0, where f 1x2 = x1x + 22 2.
11. Solve f 1x2 Ú 0, where f 1x2 = 1x + 42 2 11 - x2 .
12. Solve f 1x2 7 0, where f 1x2 = 1x - 12 1x + 32 2.
13. Solve f 1x2 … 0, where f 1x2 = - 21x + 22 1x - 22 3.
1
14. Solve f 1x2 6 0, where f 1x2 = - 1x + 42 1x - 12 3.
2
In Problems 15–18, solve the inequality by using the graph of the function.
[Hint: The graphs were drawn in Problems 7–10 of Section 5.3.]
15. Solve R 1x2 7 0, where R 1x2 =
x + 1
.
x1x + 42
16. Solve R 1x2 6 0, where R 1x2 =
x
.
1x - 12 1x + 22
17. Solve R 1x2 … 0, where R 1x2 =
3x + 3
.
2x + 4
18. Solve R 1x2 Ú 0, where R 1x2 =
2x + 4
.
x - 1
In Problems 19–48, solve each inequality algebraically.
19. 1x - 52 2 1x + 22 6 0
20. 1x - 52 1x + 22 2 7 0
21. x3 - 4x2 7 0
22. x3 + 8x2 6 0
23. 2x3 7 - 8x2
24. 3x3 6 - 15x2
25. 1x - 12 1x - 22 1x - 32 … 0
26. 1x + 12 1x + 22 1x + 32 … 0
27. x3 - 2x2 - 3x 7 0
28. x3 + 2x2 - 3x 7 0
29. x4 7 x2
30. x4 6 9x2
31. x4 7 1
32. x3 7 1
33.
34.
37.
x - 3
7 0
x + 1
1x - 22 2
x2 - 1
35.
38.
Ú 0
1x - 12 1x + 12
x
1x + 52 2
x2 - 4
36.
… 0
Ú 0
x - 4
Ú 1
2x + 4
41.
3x - 5
… 2
x + 2
42.
43.
2
1
6
x - 2
3x - 9
44.
3
5
7
x - 3
x + 1
45.
1x - 12 1x + 12
47.
Ú 0
13 - x2 3 12x + 12
x3 - 1
6 0
x - 1
x + 4
… 1
x - 2
x + 2
Ú 1
x - 4
x1x2 + 12 1x - 22
1x - 32 1x + 22
39.
40.
46.
x + 1
7 0
x - 1
48.
… 0
x2 13 + x2 1x + 42
1x + 52 1x - 12
12 - x2 3 13x - 22
x3 + 1
Ú 0
6 0
Mixed Practice
In Problems 49–60, solve each inequality algebraically.
49. 1x + 12 1x - 32 1x - 52 7 0
50. 12x - 12 1x + 22 1x + 52 6 0
52. x2 + 3x Ú 10
53.
55. 31x2 - 22 6 21x - 12 2 + x2
56. 1x - 32 1x + 22 6 x2 + 3x + 5
57. 6x - 5 6
59. x3 - 9x … 0
60. x3 - x Ú 0
58. x +
12
6 7
x
x + 1
… 2
x - 3
51. 7x - 4 Ú - 2x2
54.
x - 1
Ú -2
x + 2
6
x
In Problems 61–64, (a) graph each function by hand, and (b) solve f 1x2 Ú 0.
61. f 1x2 =
x2 + 5x - 6
x2 - 4x + 4
62. f 1x2 =
2x2 + 9x + 9
x2 - 4
63. f 1x2 =
x3 + 2x2 - 11x - 12
x2 - x - 6
64. f 1x2 =
x3 - 6x2 + 9x - 4
x2 + x - 20
374
CHAPTER 5 Polynomial and Rational Functions
Applications and Extensions
65. For what positive numbers will the cube of a number exceed
four times its square?
66. For what positive numbers will the cube of a number be less
than the number?
67. What is the domain of the function f 1x2 = 2x4 - 16?
68. What is the domain of the function f 1x2 = 2x3 - 3x2?
69. What is the domain of the function f 1x2 =
x
Ax
x
70. What is the domain of the function f 1x2 =
Ax
+
+
2
?
4
1
?
4
In Problems 71–74, determine where the graph of f is below the graph
of g by solving the inequality f 1x2 … g1x2. Graph f and g together.
71. f 1x2 = x4 - 1
g1x2 = - 2x2 + 2
73. f 1x2 = x - 4
4
g1x2 = 3x2
72. f 1x2 = x4 - 1
g1x2 = x - 1
K =
2W 1S + L2
S2
where W = weight of the jumper (pounds)
K = cord’s stiffness (pounds per foot)
L = free length of the cord (feet)
S = stretch (feet)
(a) A 150-pound person plans to jump off a ledge attached
to a cord of length 42 feet. If the stiffness of the cord is
no less than 16 pounds per foot, how much will the cord
stretch?
(b) If safety requirements will not permit the jumper to
get any closer than 3 feet to the ground, what is the
minimum height required for the ledge in part (a)?
Source: American Institute of Physics, Physics News Update,
No. 150, November 5, 1993.
78. Gravitational Force
According to Newton’s Law of
Universal Gravitation, the attractive force F between two
bodies is given by
74. f 1x2 = x4
g1x2 = 2 - x2
75. Average Cost Suppose that the daily cost C of manufacturing
bicycles is given by C 1x2 = 80x + 5000. Then the average
80x + 5000
daily cost C is given by C 1x2 =
. How many
x
bicycles must be produced each day for the average cost to
be no more than $100?
76. Average Cost See Problem 75. Suppose that the government
imposes a $1000-per-day tax on the bicycle manufacturer so
that the daily cost C of manufacturing x bicycles is now given
by C 1x2 = 80x + 6000. Now the average daily cost C is
80x + 6000
given by C 1x2 =
. How many bicycles must be
x
produced each day for the average cost to be no more than
$100?
77. Bungee Jumping Originating on Pentecost Island in the
Pacific, the practice of a person jumping from a high place
harnessed to a flexible attachment was introduced to
Western culture in 1979 by the Oxford University Dangerous
Sport Club. One important parameter to know before
attempting a bungee jump is the amount the cord will stretch
at the bottom of the fall. The stiffness of the cord is related to
the amount of stretch by the equation
F = G
m1m2
r2
where m1, m2 = the masses of the two bodies
r = distance between the two bodies
G = gravitational constant = 6.6742 * 10 - 11
newtons # meter 2 # kilogram-2
Suppose an object is traveling directly from Earth to the
moon. The mass of Earth is 5.9742 * 1024 kilograms, the
mass of the moon is 7.349 * 1022 kilograms, and the mean
distance from Earth to the moon is 384,400 kilometers. For
an object between Earth and the moon, how far from Earth
is the force on the object due to the moon greater than the
force on the object due to Earth?
Source: www.solarviews.com;en.wikipedia.org
79. Field Trip Mrs. West has decided to take her fifth grade class
to a play. The manager of the theater agreed to discount the
regular $40 price of the ticket by $0.20 for each ticket sold.
The cost of the bus, $500, will be split equally among the
students. How many students must attend to keep the cost
per student at or below $40?
Explaining Concepts: Discussion and Writing
80. Make up an inequality that has no solution. Make up one
that has exactly one solution.
81. The inequality x4 + 1 6 - 5 has no solution. Explain why.
x + 4
… 0 by
82. A student attempted to solve the inequality
x - 3
multiplying both sides of the inequality by x - 3 to
get x + 4 … 0. This led to a solution of 5x x … - 46 .
Is the student correct? Explain.
83. Write a rational
{x - 3 6 x … 5}.
inequality
whose
solution
set
Retain Your Knowledge
Problems 84–87 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your
mind so that you are better prepared for the final exam.
84. Solve: 9 - 2x … 4x + 1
86. Factor completely: 6x4y4 + 3x3y5 - 18x2y6
85. Given f 1x2 = x2 + 3x - 2, find f 1x - 22.
87. Suppose y varies directly with 2x. Write a general formula to
describe the variation if y = 2 when x = 9.
is
SECTION 5.5 The Real Zeros of a Polynomial Function
375
‘Are You Prepared?’ Answers
1
1
1. e x ` x 6 - f or a - q , - b
2
2
2. {x - 3 … x … 8} or [ - 3, 8]
2
1 1–2
0
–3
1
0
8
5.5 The Real Zeros of a Polynomial Function
PREPARING FOR THIS SECTION Before getting started, review the following:
r Evaluating Functions (Section 3.1, pp. 202–204)
r Factoring Polynomials (Chapter R, Section R.5,
pp. 49–55)
r Synthetic Division (Chapter R, Section R.6, pp. 58–61)
r Polynomial Division (Chapter R, Section R.4,
pp. 44–47)
r Solve a Quadratic Equation (Section 1.2,
pp. 92–99)
Now Work the ‘Are You Prepared?’ problems on page 386.
OBJECTIVES 1 Use the Remainder and Factor Theorems (p. 375)
2 Use Descartes’ Rule of Signs to Determine the Number of Positive and
the Number of Negative Real Zeros of a Polynomial Function (p. 378)
3 Use the Rational Zeros Theorem to List the Potential Rational Zeros of a
Polynomial Function (p. 379)
4 Find the Real Zeros of a Polynomial Function (p. 380)
5 Solve Polynomial Equations (p. 382)
6 Use the Theorem for Bounds on Zeros (p. 383)
7 Use the Intermediate Value Theorem (p. 384)
In Section 5.1, we were able to identify the real zeros of a polynomial function
because either the polynomial function was in factored form or it could be easily
factored. But how do we find the real zeros of a polynomial function if it is not
factored or cannot be easily factored?
Recall that if r is a real zero of a polynomial function f, then f1r2 = 0, r is an
x-intercept of the graph of f, x - r is a factor of f, and r is a solution of the equation
f1x2 = 0. For example, if x - 4 is a factor of f, then 4 is a real zero of f and 4 is
a solution to the equation f1x2 = 0. For polynomial functions, we have seen the
importance of the real zeros for graphing. In most cases, however, the real zeros of a
polynomial function are difficult to find using algebraic methods. No nice formulas
like the quadratic formula are available to help us find zeros for polynomials of
degree 3 or higher. Formulas do exist for solving any third- or fourth-degree polynomial
equation, but they are somewhat complicated. No general formulas exist for
polynomial equations of degree 5 or higher. Refer to the Historical Feature at the
end of this section for more information.
1 Use the Remainder and Factor Theorems
When one polynomial (the dividend) is divided by another (the divisor), a quotient
polynomial and a remainder are obtained, the remainder being either the zero
polynomial or a polynomial whose degree is less than the degree of the divisor. To
check, verify that
1Quotient2 1Divisor2 + Remainder = Dividend
This checking routine is the basis for a famous theorem called the division
algorithm* for polynomials, which we now state without proof.
*A systematic process in which certain steps are repeated a finite number of times is called an algorithm.
For example, long division is an algorithm.
376
CHAPTER 5 Polynomial and Rational Functions
THEOREM
Division Algorithm for Polynomials
If f1x2 and g1x2 denote polynomial functions and if g1x2 is a polynomial
whose degree is greater than zero, then there are unique polynomial functions
q 1x2 and r 1x2 such that
r 1x2
f1x2
= q 1x2 +
g1x2
g1x2
or f1x2 = q 1x2g1x2 + r 1x2
c
dividend
c
quotient
c
(1)
c
divisor remainder
where r 1x2 is either the zero polynomial or a polynomial of degree less than
that of g1x2 .
In equation (1), f1x2 is the dividend, g1x2 is the divisor, q 1x2 is the quotient,
and r 1x2 is the remainder.
If the divisor g1x2 is a first-degree polynomial of the form
g1x2 = x - c
c a real number
then the remainder r 1x2 is either the zero polynomial or a polynomial of degree 0.
As a result, for such divisors, the remainder is some number, say R, and we may write
f1x2 = 1x - c2q 1x2 + R
(2)
This equation is an identity in x and is true for all real numbers x. Suppose that
x = c. Then equation (2) becomes
f 1c2 = 1c - c2q 1c2 + R
f1c2 = R
Substitute f1c2 for R in equation (2) to obtain
f1x2 = 1x - c2q 1x2 + f1c2
(3)
which proves the Remainder Theorem.
REMAINDER THEOREM
EX AMPLE 1
Let f be a polynomial function. If f 1x2 is divided by x - c, then the remainder
is f1c2 .
Using the Remainder Theorem
Find the remainder when f1x2 = x3 - 4x2 - 5 is divided by
(a) x - 3
Solution
(b) x + 2
(a) Either long division or synthetic division could be used, but it is easier to use the
Remainder Theorem, which says that the remainder is f132 .
f 132 = 132 3 - 4132 2 - 5 = 27 - 36 - 5 = - 14
The remainder is - 14.
(b) To find the remainder when f1x2 is divided by x + 2 = x - ( - 2), find f( - 2).
f1 - 22 = 1 - 22 3 - 41 - 22 2 - 5 = - 8 - 16 - 5 = - 29
The remainder is - 29.
r
Compare the method used in Example 1(a) with the method used in Example 1 of
Chapter R, Section R.6. Which method do you prefer?
SECTION 5.5 The Real Zeros of a Polynomial Function
377
COMMENT A graphing utility provides another way to find the value of a function using the eVALUEate
feature. Consult your manual for details. Then check the results of Example 1.
■
An important and useful consequence of the Remainder Theorem is the Factor
Theorem.
FACTOR THEOREM
Let f be a polynomial function. Then x - c is a factor of f1x2 if and only if
f1c2 = 0.
The Factor Theorem actually consists of two separate statements:
1. If f1c2 = 0, then x - c is a factor of f 1x2 .
2. If x - c is a factor of f1x2 , then f 1c2 = 0.
The proof requires two parts.
Proof
1. Suppose that f1c2 = 0. Then, by equation (3), we have
f 1x2 = 1x - c2q 1x2
for some polynomial q 1x2. That is, x - c is a factor of f 1x2.
2. Suppose that x - c is a factor of f1x2. Then there is a polynomial function q
such that
f 1x2 = 1x - c2 q 1x2
Replacing x by c, we find that
f 1c2 = 1c - c2q 1c2 = 0 # q 1c2 = 0
This completes the proof.
■
One use of the Factor Theorem is to determine whether a polynomial has a
particular factor.
EXAM PL E 2
Using the Factor Theorem
Use the Factor Theorem to determine whether the function
f1x2 = 2x3 - x2 + 2x - 3
has the factor
(a) x - 1
Solution
(b) x + 2
The Factor Theorem states that if f1c2 = 0, then x - c is a factor.
(a) Because x - 1 is of the form x - c with c = 1, find the value of f112. We choose
to use substitution.
f112 = 2112 3 - 112 2 + 2112 - 3 = 2 - 1 + 2 - 3 = 0
By the Factor Theorem, x - 1 is a factor of f1x2.
(b) To test the factor x + 2, first write it in the form x - c. Since x + 2 = x - 1 - 22 ,
find the value of f1 - 22. We choose to use synthetic division.
- 2) 2
2
-1 2
- 4 10
- 5 12
-3
- 24
- 27
378
CHAPTER 5 Polynomial and Rational Functions
Because f1 - 22 = - 27 ≠ 0, conclude from the Factor Theorem that
x - 1 - 22 = x + 2 is not a factor of f1x2.
r
Now Work
PROBLEM
11
In Example 2(a), x - 1 was found to be a factor of f. To write f in factored
form, use long division or synthetic division.
1) 2
2
-1 2
2 1
1 3
-3
3
0
The quotient is q 1x2 = 2x + x + 3 with a remainder of 0, as expected. Write f in
factored form as
2
f 1x2 = 2x3 - x2 + 2x - 3 = 1x - 12 12x2 + x + 32
The next theorem concerns the number of real zeros that a polynomial function
may have. In counting the zeros of a polynomial, count each zero as many times as
its multiplicity.
THEOREM
Number of Real Zeros
A polynomial function cannot have more real zeros than its degree.
Proof The proof is based on the Factor Theorem. If r is a real zero of a polynomial
function f, then f1r2 = 0, and x - r is a factor of f1x2. Each real zero corresponds
to a factor of degree 1. Because f cannot have more first-degree factors than its
degree, the result follows.
■
2 Use Descartes’ Rule of Signs to Determine the Number
of Positive and the Number of Negative Real Zeros
of a Polynomial Function
Descartes’ Rule of Signs provides information about the number and location of the
real zeros of a polynomial function written in standard form (omitting terms with a
0 coefficient). It utilizes the number of variations in the sign of the coefficients of
f1x2 and f1 - x2.
For example, the following polynomial function has two variations in the signs
of the coefficients.
f1x2 = - 3x7 + 4x4 + 3x2 - 2x - 1
- to +
+ to -
Replacing x by - x gives
f1 - x2 = - 31 - x2 7 + 41 - x2 4 + 31 - x2 2 - 21 - x2 - 1
= 3x7 + 4x4 + 3x2 + 2x - 1
+ to -
which has one variation in sign.
THEOREM
Descartes’ Rule of Signs
Let f denote a polynomial function written in standard form.
The number of positive real zeros of f either equals the number of variations
in the sign of the nonzero coefficients of f1x2 or else equals that number less
an even integer.
The number of negative real zeros of f either equals the number of variations
in the sign of the nonzero coefficients of f1 - x2 or else equals that number less
an even integer.
We shall not prove Descartes’ Rule of Signs. Let’s see how it is used.
SECTION 5.5 The Real Zeros of a Polynomial Function
EXAMPL E 3
379
Using the Number of Real Zeros Theorem
and Descartes’ Rule of Signs
Discuss the real zeros of f1x2 = 3x7 - 4x4 + 3x3 + 2x2 - x - 3.
Solution
Because the polynomial is of degree 7 , by the Number of Real Zeros Theorem
there are at most seven real zeros. Since there are three variations in the sign of the
nonzero coefficients of f1x2, by Descartes’ Rule of Signs we expect either three
positive real zeros or one positive real zero. To continue, look at f 1 - x2 .
f 1 - x2 = - 3x7 - 4x4 - 3x3 + 2x2 + x - 3
There are two variations in sign, so we expect either two negative real zeros or no
negative real zeros. Equivalently, we now know that the graph of f has either three
positive x-intercepts or one positive x-intercept and two negative x intercepts or no
negative x-intercepts.
r
Now Work
PROBLEM
21
3 Use the Rational Zeros Theorem to List the Potential Rational
Zeros of a Polynomial Function
The next result, called the Rational Zeros Theorem, provides information about the
rational zeros of a polynomial with integer coefficients.
THEOREM
Rational Zeros Theorem
Let f be a polynomial function of degree 1 or higher of the form
f1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0
an ≠ 0 a0 ≠ 0
p
, in lowest terms, is a rational zero
q
of f, then p must be a factor of a0 , and q must be a factor of an.
where each coefficient is an integer. If
EXAMPL E 4
Listing Potential Rational Zeros
List the potential rational zeros of
f1x2 = 2x3 + 11x2 - 7x - 6
Solution
Because f has integer coefficients, the Rational Zeros Theorem may be used. First,
list all the integers p that are factors of the constant term a0 = - 6 and all the
integers q that are factors of the leading coefficient a3 = 2.
p:
{1, {2, {3, {6
Factors of - 6
q:
{1, {2
Factors of 2
p
Now form all possible ratios .
q
p
:
q
In Words
For the polynomial function
f (x) = 2x 3 + 11x 2 - 7x - 6,
we know 5 is not a zero, because
5 is not in the list of potential
rational zeros. However, - 1 may
or may not be a zero.
1
2
3
6
1
2
3
6
{ ,{ ,{ ,{ ,{ ,{ ,{ ,{
1
1
1
1
2
2
2
2
which simplify to
p
:
q
1
3
{1, {2, {3, {6, { , {
2
2
If f has a rational zero, it will be found in this list, which contains 12 possibilities.
Now Work
PROBLEM
33
r
380
CHAPTER 5 Polynomial and Rational Functions
Be sure that you understand what the Rational Zeros Theorem says: For a
polynomial with integer coefficients, if there is a rational zero, it is one of those
listed. It may be the case that the function does not have any rational zeros.
Long division, synthetic division, or substitution can be used to test each
potential rational zero to determine whether it is indeed a zero. To make the work
easier, integers are usually tested first.
4 Find the Real Zeros of a Polynomial Function
EX AMPLE 5
How to Find the Real Zeros of a Polynomial Function
Find the real zeros of the polynomial function f 1x2 = 2x3 + 11x2 - 7x - 6. Write f
in factored form.
Step-by-Step Solution
Step 1: Use the degree of the
polynomial to determine the
maximum number of zeros.
Since f is a polynomial of degree 3, there are at most three real zeros.
Step 2: Use Descartes’ Rule of
Signs to determine the possible
number of positive zeros and
negative zeros.
By Descartes’ Rule of Signs, there is one positive real zero. Also, because
Step 3: If the polynomial has integer
coefficients, use the Rational Zeros
Theorem to identify those rational
numbers that potentially can be
zeros. Use the Factor Theorem to
determine whether each potential
rational zero is a zero. If it is, use
synthetic division or long division
to factor the polynomial function.
Repeat Step 3 until all the rational
zeros of the polynomial function
have been identified.
List the potential rational zeros obtained in Example 4:
f 1 - x2 = - 2x3 + 11x2 + 7x - 6
there are two negative zeros or no negative zeros.
1
3
{1, {2, {3, {6, { , {
2
2
From our list of potential rational zeros, we will test 6 to determine whether it is
a zero of f. Because f 162 = 780 ≠ 0, we know that 6 is not a zero of f. Now, let’s
test whether - 6 is a zero. Because f 1 - 62 = 0, we know that - 6 is a zero and
x - 1 - 62 = x + 6 is a factor of f. Use long division or synthetic division to factor f.
(We will not show the division here, but you are encouraged to verify the results
shown.) After dividing f by x + 6, the quotient is 2x2 - x - 1, so
f 1x2 = 2x3 + 11x2 - 7x - 6
= (x + 6)(2x2 - x - 1)
Now any solution of the equation 2x2 - x - 1 = 0 will be a zero of f. The equation
2x2 - x - 1 = 0 is called a depressed equation of f. Because any solution to the
equation 2x2 - x - 1 = 0 is a zero of f, work with the depressed equation to find
the remaining zeros of f.
The depressed equation 2x2 - x - 1 = 0 is a quadratic equation with
discriminant b2 - 4ac = 1 - 12 2 - 4122 1 - 12 = 9 7 0. The equation has two real
solutions, which can be found by factoring.
2x2 - x - 1 = 12x + 12 1x - 12 = 0
2x + 1 = 0
or
x - 1 = 0
1
x = or
x = 1
2
1
The zeros of f are - 6, - , and 1.
2
Factor f completely as follows:
f1x2 = 2x3 + 11x2 - 7x - 6 = 1x + 62 12x2 - x - 12
= 1x + 62 12x + 12 1x - 12
Notice that all three zeros of f are in the list of potential rational zeros. Also note
that the zeros confirm what was expected from Descartes’ Rule of Signs.
r
SECTION 5.5 The Real Zeros of a Polynomial Function
381
SUMMARY
Steps for Finding the Real Zeros of a Polynomial Function
STEP 1: Use the degree of the polynomial to determine the maximum number of real zeros.
STEP 2: Use Descartes’ Rule of Signs to determine the possible number of positive zeros and negative zeros.
STEP 3: (a) If the polynomial has integer coefficients, use the Rational Zeros Theorem to identify those rational
numbers that potentially could be zeros.
(b) Use substitution, synthetic division, or long division to test each potential rational zero. Each time that
a zero (and thus a factor) is found, repeat Step 3 on the depressed equation.
In attempting to find the zeros, remember to use (if possible) the factoring techniques that you already
know (special products, factoring by grouping, and so on).
EXAMPL E 6
Solution
Finding the Real Zeros of a Polynomial Function
Find the real zeros of f 1x2 = x5 - 7x4 + 19x3 - 37x2 + 60x - 36. Write f in
factored form.
STEP 1: Because f is a polynomial of degree 5, there are at most five real zeros.
STEP 2: By Descartes’ Rule of Signs, there are five, three, or one positive zeros.
There are no negative zeros because
f1 - x2 = - x5 - 7x4 - 19x3 - 37x2 - 60x - 36
has no sign variation.
STEP 3: Because the leading coefficient a5 = 1 and there are no negative zeros, the
potential rational zeros are limited to the positive integers 1, 2, 3, 4, 6, 9, 12, 18, and
36 (the positive factors of the constant term, 36). Test the potential rational zero 1
first, using synthetic division.
1) 1
1
- 7 19 - 37 60 - 36
1 -6
13 - 24
36
- 6 13 - 24 36
0
The remainder is f 112 = 0, so 1 is a zero and x - 1 is a factor of f. Use the entries
in the bottom row of the synthetic division to begin factoring f.
f1x2 = x5 - 7x4 + 19x3 - 37x2 + 60x - 36
= 1x - 12 1x4 - 6x3 + 13x2 - 24x + 362
Continue the process using the first depressed equation:
q1 1x2 = x4 - 6x3 + 13x2 - 24x + 36 = 0
REPEAT STEP 3: The potential rational zeros of q1 are still 1, 2, 3, 4, 6, 9, 12, 18, and
36. Test 1 again, since it may be a repeated zero of f.
1 ) 1 - 6 13 - 24 36
1 -5
8 - 16
1 -5
8 - 16 20
Since the remainder is 20, 1 is not a repeated zero. Try 2 next.
2) 1
- 6 13 - 24 36
2 - 8 10 - 28
1 -4
5 - 14
8
Since the remainder is 8, 2 is not a zero. Try 3 next.
3 ) 1 - 6 13 - 24 36
3 - 9 12 - 36
1 -3
4 - 12
0
382
CHAPTER 5 Polynomial and Rational Functions
The remainder is f132 = 0, so 3 is a zero and x - 3 is a factor of f. Use the bottom
row of the synthetic division to continue the factoring of f.
f1x2 = x5 - 7x4 + 19x3 - 37x2 + 60x - 36
= 1x - 12 1x - 32 1x3 - 3x2 + 4x - 122
The remaining zeros satisfy the new depressed equation
q2 1x2 = x3 - 3x2 + 4x - 12 = 0
Notice that q2 1x2 can be factored by grouping. (Alternatively, Step 3 could be
repeated to again check the potential rational zero 3. The potential rational zeros
1 or 2 would no longer be checked, because they have already been eliminated from
further consideration.) Then
x3 - 3x2 + 4x - 12
x 1x - 32 + 41x - 32
1x2 + 42 1x - 32
x2 + 4 = 0 or x - 3
x
2
=
=
=
=
=
0
0
0
0
3
Since x2 + 4 = 0 has no real solutions, the real zeros of f are 1 and 3, with 3 being a
repeated zero of multiplicity 2. The factored form of f is
f1x2 = x5 - 7x4 + 19x3 - 37x2 + 60x - 36
= 1x - 12 1x - 32 2 1x2 + 42
Now Work
PROBLEM
r
45
5 Solve Polynomial Equations
EX AMPLE 7
Solving a Polynomial Equation
Find the real solutions of the equation: x5 - 7x4 + 19x3 - 37x2 + 60x - 36 = 0
Solution
The real solutions of this equation are the real zeros of the polynomial function
f1x2 = x5 - 7x4 + 19x3 - 37x2 + 60x - 36
Using the result of Example 6, the real zeros of f are 1 and 3. The real solutions of the
equation x5 - 7x4 + 19x3 - 37x2 + 60x - 36 = 0 are 1 and 3.
r
Now Work
PROBLEM
57
In Example 6, the quadratic factor x2 + 4 that appears in the factored form of f is
called irreducible, because the polynomial x2 + 4 cannot be factored over the real
numbers. In general, a quadratic factor ax2 + bx + c is irreducible if it cannot be
factored over the real numbers—that is, if it is prime over the real numbers.
Refer to Examples 5 and 6. The polynomial function of Example 5 has three
real zeros, and its factored form contains three linear factors. The polynomial
function of Example 6 has two distinct real zeros, and its factored form contains two
distinct linear factors and one irreducible quadratic factor.
THEOREM
Every polynomial function (with real coefficients) can be uniquely factored into
a product of linear factors and/or irreducible quadratic factors.
We prove this result in Section 5.6, and in fact, we shall draw several additional
conclusions about the zeros of a polynomial function. One conclusion is worth
noting now. If a polynomial with real coefficients is of odd degree, it must contain at
least one linear factor. (Do you see why? Consider the end behavior of polynomial
functions of odd degree.) This means that it must have at least one real zero.
SECTION 5.5 The Real Zeros of a Polynomial Function
THEOREM
383
A polynomial function (with real coefficients) of odd degree has at least one
real zero.
6 Use the Theorem for Bounds on Zeros
The work involved in finding the zeros of a polynomial function can be reduced
somewhat if upper and lower bounds to the zeros can be found. A number M is an
upper bound to the zeros of a polynomial f if no zero of f is greater than M. The
number m is a lower bound if no zero of f is less than m. Accordingly, if m is a lower
bound and M is an upper bound to the zeros of a polynomial function f, then
COMMENT The bounds on the zeros
of a polynomial provide good choices for
setting Xmin and Xmax of the viewing
rectangle. With these choices, all the
x-intercepts of the graph can be seen. ■
THEOREM
m … any zero of f … M
For polynomials with integer coefficients, knowing the values of a lower bound m
and an upper bound M may enable you to eliminate some potential rational
zeros—that is, any zeros outside of the interval [m, M].
Bounds on Zeros
Let f denote a polynomial function whose leading coefficient is positive.
r If M 7 0 is a real number and if the third row in the process of synthetic
division of f by x - M contains only numbers that are positive or zero,
then M is an upper bound to the zeros of f.
r If m 6 0 is a real number and if the third row in the process of synthetic
division of f by x - m contains numbers that alternate positive (or 0) and
negative (or 0), then m is a lower bound to the zeros of f.
NOTE When finding a lower bound,
remember that a 0 can be treated as
either positive or negative, but not both.
For example, 3, 0, 5 would be considered
to alternate sign, whereas 3, 0, - 5
would not.
■
Proof (Outline) We give only an outline of the proof of the first part of the theorem.
Suppose that M is a positive real number, and the third row in the process of
synthetic division of the polynomial f by x - M contains only numbers that are
positive or 0. Then there are a quotient q and a remainder R such that
f1x2 = 1x - M2q 1x2 + R
where the coefficients of q 1x2 are positive or 0 and the remainder R Ú 0. Then, for
any x 7 M, we must have x - M 7 0, q 1x2 7 0, and R Ú 0, so that f1x2 7 0.
That is, there is no zero of f larger than M. The proof of the second part follows similar
reasoning.
■
In finding bounds, it is preferable to find the smallest upper bound and largest
lower bound. This will require repeated synthetic division until a desired pattern
is observed. For simplicity, we will consider only potential rational zeros that are
integers. If a bound is not found using these values, continue checking positive
and/or negative integers until you find both an upper and a lower bound.
EXAMPL E 8
Solution
Finding Upper and Lower Bounds of Zeros
For the polynomial function f 1x2 = 2x3 + 11x2 - 7x - 6, use the Bounds on Zeros
Theorem to find integer upper and lower bounds to the zeros of f.
3
1
From Example 4, the potential rational zeros of f are {1, {2, {3, {6, { , { .
2
2
To find an upper bound, start with the smallest positive integer that is a potential
rational zero, which is 1. Continue checking 2, 3, and 6 (and then subsequent positive
integers), if necessary, until an upper bound is found. To find a lower bound, start with
the largest negative integer that is a potential rational zero, which is - 1. Continue
checking - 2, - 3, and - 6 (and then subsequent negative integers), if necessary, until
a lower bound is found. Table 16 summarizes the results of doing repeated synthetic
384
CHAPTER 5 Polynomial and Rational Functions
divisions by showing only the third row of each division. For example, the first row
of the table shows the result of dividing f 1x2 by x - 1.
1 ) 2 11 - 7
2 13
2 13
6
-6
6
0
Table 16 Synthetic Division Summary
Upper
bound
NOTE Keep track of any zeros that are
found when looking for bounds.
■
Lower
bound
r
Coefficients of q(x)
1
2
13
6
0
-1
2
9
- 16
10
-2
2
7
- 21
36
-3
2
5
- 22
60
-6
2
-1
-1
0
-7
2
-3
14
- 104
Remainder
All nonnegative
Alternating
Signs
For r = 1, the third row of synthetic division contains only numbers that are positive
or 0, so we know there are no zeros greater than 1. Since the third row of synthetic
division for r = - 7 results in alternating positive (or 0) and negative (or 0) values,
we know that - 7 is a lower bound. There are no zeros less than - 7. Notice that in
looking for bounds, two zeros were discovered. These zeros are 1 and - 6.
r
y
Now Work
y 5 f (x )
f(b)
f (b )
Zero
a
b
x
PROBLEM
69
If the leading coefficient of f is negative, the upper and lower bounds can still be
found by first multiplying the polynomial by - 1. Since - f1x2 = 1 - 12 f1x2 , the
zeros of - f1x2 are the same as the zeros of f1x2 .
f (a)
7 Use the Intermediate Value Theorem
f(a)
Figure 45 If f(a) 6 0 and f(b) 7 0,
and if f is continuous, there is a zero
between a and b.
THEOREM
The next result, called the Intermediate Value Theorem, is based on the fact that the
graph of a polynomial function is continuous; that is, it contains no “holes” or “gaps.”
Although the proof of this result requires advanced methods in calculus, it is easy to
“see” why the result is true. Look at Figure 45.
Intermediate Value Theorem
Let f denote a polynomial function. If a 6 b and if f1a2 and f1b2 are of opposite
sign, there is at least one real zero of f between a and b.
EX AMPLE 9
Using the Intermediate Value Theorem to Locate a Real Zero
Show that f1x2 = x5 - x3 - 1 has a zero between 1 and 2.
Solution
Evaluate f at 1 and at 2.
f112 = - 1 and f122 = 23
Because f112 6 0 and f 122 7 0, it follows from the Intermediate Value Theorem
that the polynomial function f has at least one zero between 1 and 2.
r
Now Work
PROBLEM
79
SECTION 5.5 The Real Zeros of a Polynomial Function
385
Let’s look at the polynomial f of Example 9 more closely. Based on Descartes’ Rule
of Signs, f has exactly one positive real zero. Based on the